1281 lines
37 KiB
Java
1281 lines
37 KiB
Java
<!-- GFM-TOC -->
|
||
* [BFS](#bfs)
|
||
* [1. 计算在网格中从原点到特定点的最短路径长度](#1-计算在网格中从原点到特定点的最短路径长度)
|
||
* [2. 组成整数的最小平方数数量](#2-组成整数的最小平方数数量)
|
||
* [3. 最短单词路径](#3-最短单词路径)
|
||
* [DFS](#dfs)
|
||
* [1. 查找最大的连通面积](#1-查找最大的连通面积)
|
||
* [2. 矩阵中的连通分量数目](#2-矩阵中的连通分量数目)
|
||
* [3. 好友关系的连通分量数目](#3-好友关系的连通分量数目)
|
||
* [4. 填充封闭区域](#4-填充封闭区域)
|
||
* [5. 能到达的太平洋和大西洋的区域](#5-能到达的太平洋和大西洋的区域)
|
||
* [Backtracking](#backtracking)
|
||
* [1. 数字键盘组合](#1-数字键盘组合)
|
||
* [2. IP 地址划分](#2-ip-地址划分)
|
||
* [3. 在矩阵中寻找字符串](#3-在矩阵中寻找字符串)
|
||
* [4. 输出二叉树中所有从根到叶子的路径](#4-输出二叉树中所有从根到叶子的路径)
|
||
* [5. 排列](#5-排列)
|
||
* [6. 含有相同元素求排列](#6-含有相同元素求排列)
|
||
* [7. 组合](#7-组合)
|
||
* [8. 组合求和](#8-组合求和)
|
||
* [9. 含有相同元素的组合求和](#9-含有相同元素的组合求和)
|
||
* [10. 1-9 数字的组合求和](#10-1-9-数字的组合求和)
|
||
* [11. 子集](#11-子集)
|
||
* [12. 含有相同元素求子集](#12-含有相同元素求子集)
|
||
* [13. 分割字符串使得每个部分都是回文数](#13-分割字符串使得每个部分都是回文数)
|
||
* [14. 数独](#14-数独)
|
||
* [15. N 皇后](#15-n-皇后)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
深度优先搜索和广度优先搜索广泛运用于树和图中，但是它们的应用远远不止如此。
|
||
|
||
# BFS
|
||
|
||
<div align="center"> <img src="pics/95903878-725b-4ed9-bded-bc4aae0792a9.jpg"/> </div><br>
|
||
|
||
广度优先搜索一层一层地进行遍历，每层遍历都以上一层遍历的结果作为起点，遍历一个距离能访问到的所有节点。需要注意的是，遍历过的节点不能再次被遍历。
|
||
|
||
第一层：
|
||
|
||
- 0 -> {6,2,1,5}
|
||
|
||
第二层：
|
||
|
||
- 6 -> {4}
|
||
- 2 -> {}
|
||
- 1 -> {}
|
||
- 5 -> {3}
|
||
|
||
第三层：
|
||
|
||
- 4 -> {}
|
||
- 3 -> {}
|
||
|
||
每一层遍历的节点都与根节点距离相同。设 d<sub>i</sub> 表示第 i 个节点与根节点的距离，推导出一个结论：对于先遍历的节点 i 与后遍历的节点 j，有 d<sub>i</sub> <= d<sub>j</sub>。利用这个结论，可以求解最短路径等 **最优解** 问题：第一次遍历到目的节点，其所经过的路径为最短路径。应该注意的是，使用 BFS 只能求解无权图的最短路径，无权图是指从一个节点到另一个节点的代价都记为 1。
|
||
|
||
在程序实现 BFS 时需要考虑以下问题：
|
||
|
||
- 队列：用来存储每一轮遍历得到的节点；
|
||
- 标记：对于遍历过的节点，应该将它标记，防止重复遍历。
|
||
|
||
## 1. 计算在网格中从原点到特定点的最短路径长度
|
||
|
||
```html
|
||
[[1,1,0,1],
|
||
[1,0,1,0],
|
||
[1,1,1,1],
|
||
[1,0,1,1]]
|
||
```
|
||
|
||
题目描述：1 表示可以经过某个位置，求解从 (0, 0) 位置到 (tr, tc) 位置的最短路径长度。
|
||
|
||
```java
|
||
public int minPathLength(int[][] grids, int tr, int tc) {
|
||
final int[][] direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
|
||
final int m = grids.length, n = grids[0].length;
|
||
Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
|
||
queue.add(new Pair<>(0, 0));
|
||
int pathLength = 0;
|
||
while (!queue.isEmpty()) {
|
||
int size = queue.size();
|
||
pathLength++;
|
||
while (size-- > 0) {
|
||
Pair<Integer, Integer> cur = queue.poll();
|
||
int cr = cur.getKey(), cc = cur.getValue();
|
||
grids[cr][cc] = 0; // 标记
|
||
for (int[] d : direction) {
|
||
int nr = cr + d[0], nc = cc + d[1];
|
||
if (nr < 0 || nr >= m || nc < 0 || nc >= n || grids[nr][nc] == 0) {
|
||
continue;
|
||
}
|
||
if (nr == tr && nc == tc) {
|
||
return pathLength;
|
||
}
|
||
queue.add(new Pair<>(nr, nc));
|
||
}
|
||
}
|
||
}
|
||
return -1;
|
||
}
|
||
```
|
||
|
||
## 2. 组成整数的最小平方数数量
|
||
|
||
[279. Perfect Squares (Medium)](https://leetcode.com/problems/perfect-squares/description/)
|
||
|
||
```html
|
||
For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
|
||
```
|
||
|
||
可以将每个整数看成图中的一个节点，如果两个整数之差为一个平方数，那么这两个整数所在的节点就有一条边。
|
||
|
||
要求解最小的平方数数量，就是求解从节点 n 到节点 0 的最短路径。
|
||
|
||
本题也可以用动态规划求解，在之后动态规划部分中会再次出现。
|
||
|
||
```java
|
||
public int numSquares(int n) {
|
||
List<Integer> squares = generateSquares(n);
|
||
Queue<Integer> queue = new LinkedList<>();
|
||
boolean[] marked = new boolean[n + 1];
|
||
queue.add(n);
|
||
marked[n] = true;
|
||
int level = 0;
|
||
while (!queue.isEmpty()) {
|
||
int size = queue.size();
|
||
level++;
|
||
while (size-- > 0) {
|
||
int cur = queue.poll();
|
||
for (int s : squares) {
|
||
int next = cur - s;
|
||
if (next < 0) {
|
||
break;
|
||
}
|
||
if (next == 0) {
|
||
return level;
|
||
}
|
||
if (marked[next]) {
|
||
continue;
|
||
}
|
||
marked[next] = true;
|
||
queue.add(next);
|
||
}
|
||
}
|
||
}
|
||
return n;
|
||
}
|
||
|
||
/**
|
||
* 生成小于 n 的平方数序列
|
||
* @return 1,4,9,...
|
||
*/
|
||
private List<Integer> generateSquares(int n) {
|
||
List<Integer> squares = new ArrayList<>();
|
||
int square = 1;
|
||
int diff = 3;
|
||
while (square <= n) {
|
||
squares.add(square);
|
||
square += diff;
|
||
diff += 2;
|
||
}
|
||
return squares;
|
||
}
|
||
```
|
||
|
||
## 3. 最短单词路径
|
||
|
||
[127. Word Ladder (Medium)](https://leetcode.com/problems/word-ladder/description/)
|
||
|
||
```html
|
||
Input:
|
||
beginWord = "hit",
|
||
endWord = "cog",
|
||
wordList = ["hot","dot","dog","lot","log","cog"]
|
||
|
||
Output: 5
|
||
|
||
Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
|
||
return its length 5.
|
||
```
|
||
|
||
```html
|
||
Input:
|
||
beginWord = "hit"
|
||
endWord = "cog"
|
||
wordList = ["hot","dot","dog","lot","log"]
|
||
|
||
Output: 0
|
||
|
||
Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
|
||
```
|
||
|
||
题目描述：找出一条从 beginWord 到 endWord 的最短路径，每次移动规定为改变一个字符，并且改变之后的字符串必须在 wordList 中。
|
||
|
||
```java
|
||
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
|
||
wordList.add(beginWord);
|
||
int N = wordList.size();
|
||
int start = N - 1;
|
||
int end = 0;
|
||
while (end < N && !wordList.get(end).equals(endWord)) {
|
||
end++;
|
||
}
|
||
if (end == N) {
|
||
return 0;
|
||
}
|
||
List<Integer>[] graphic = buildGraphic(wordList);
|
||
return getShortestPath(graphic, start, end);
|
||
}
|
||
|
||
private List<Integer>[] buildGraphic(List<String> wordList) {
|
||
int N = wordList.size();
|
||
List<Integer>[] graphic = new List[N];
|
||
for (int i = 0; i < N; i++) {
|
||
graphic[i] = new ArrayList<>();
|
||
for (int j = 0; j < N; j++) {
|
||
if (isConnect(wordList.get(i), wordList.get(j))) {
|
||
graphic[i].add(j);
|
||
}
|
||
}
|
||
}
|
||
return graphic;
|
||
}
|
||
|
||
private boolean isConnect(String s1, String s2) {
|
||
int diffCnt = 0;
|
||
for (int i = 0; i < s1.length() && diffCnt <= 1; i++) {
|
||
if (s1.charAt(i) != s2.charAt(i)) {
|
||
diffCnt++;
|
||
}
|
||
}
|
||
return diffCnt == 1;
|
||
}
|
||
|
||
private int getShortestPath(List<Integer>[] graphic, int start, int end) {
|
||
Queue<Integer> queue = new LinkedList<>();
|
||
boolean[] marked = new boolean[graphic.length];
|
||
queue.add(start);
|
||
marked[start] = true;
|
||
int path = 1;
|
||
while (!queue.isEmpty()) {
|
||
int size = queue.size();
|
||
path++;
|
||
while (size-- > 0) {
|
||
int cur = queue.poll();
|
||
for (int next : graphic[cur]) {
|
||
if (next == end) {
|
||
return path;
|
||
}
|
||
if (marked[next]) {
|
||
continue;
|
||
}
|
||
marked[next] = true;
|
||
queue.add(next);
|
||
}
|
||
}
|
||
}
|
||
return 0;
|
||
}
|
||
```
|
||
|
||
# DFS
|
||
|
||
<div align="center"> <img src="pics/74dc31eb-6baa-47ea-ab1c-d27a0ca35093.png"/> </div><br>
|
||
|
||
广度优先搜索一层一层遍历，每一层得到的所有新节点，要用队列存储起来以备下一层遍历的时候再遍历。
|
||
|
||
而深度优先搜索在得到一个新节点时立即对新节点进行遍历：从节点 0 出发开始遍历，得到到新节点 6 时，立马对新节点 6 进行遍历，得到新节点 4；如此反复以这种方式遍历新节点，直到没有新节点了，此时返回。返回到根节点 0 的情况是，继续对根节点 0 进行遍历，得到新节点 2，然后继续以上步骤。
|
||
|
||
从一个节点出发，使用 DFS 对一个图进行遍历时，能够遍历到的节点都是从初始节点可达的，DFS 常用来求解这种 **可达性** 问题。
|
||
|
||
在程序实现 DFS 时需要考虑以下问题：
|
||
|
||
- 栈：用栈来保存当前节点信息，当遍历新节点返回时能够继续遍历当前节点。可以使用递归栈。
|
||
- 标记：和 BFS 一样同样需要对已经遍历过的节点进行标记。
|
||
|
||
## 1. 查找最大的连通面积
|
||
|
||
[695. Max Area of Island (Medium)](https://leetcode.com/problems/max-area-of-island/description/)
|
||
|
||
```html
|
||
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
|
||
[0,0,0,0,0,0,0,1,1,1,0,0,0],
|
||
[0,1,1,0,1,0,0,0,0,0,0,0,0],
|
||
[0,1,0,0,1,1,0,0,1,0,1,0,0],
|
||
[0,1,0,0,1,1,0,0,1,1,1,0,0],
|
||
[0,0,0,0,0,0,0,0,0,0,1,0,0],
|
||
[0,0,0,0,0,0,0,1,1,1,0,0,0],
|
||
[0,0,0,0,0,0,0,1,1,0,0,0,0]]
|
||
```
|
||
|
||
```java
|
||
private int m, n;
|
||
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
|
||
|
||
public int maxAreaOfIsland(int[][] grid) {
|
||
if (grid == null || grid.length == 0) {
|
||
return 0;
|
||
}
|
||
m = grid.length;
|
||
n = grid[0].length;
|
||
int maxArea = 0;
|
||
for (int i = 0; i < m; i++) {
|
||
for (int j = 0; j < n; j++) {
|
||
maxArea = Math.max(maxArea, dfs(grid, i, j));
|
||
}
|
||
}
|
||
return maxArea;
|
||
}
|
||
|
||
private int dfs(int[][] grid, int r, int c) {
|
||
if (r < 0 || r >= m || c < 0 || c >= n || grid[r][c] == 0) {
|
||
return 0;
|
||
}
|
||
grid[r][c] = 0;
|
||
int area = 1;
|
||
for (int[] d : direction) {
|
||
area += dfs(grid, r + d[0], c + d[1]);
|
||
}
|
||
return area;
|
||
}
|
||
```
|
||
|
||
## 2. 矩阵中的连通分量数目
|
||
|
||
[200. Number of Islands (Medium)](https://leetcode.com/problems/number-of-islands/description/)
|
||
|
||
```html
|
||
Input:
|
||
11000
|
||
11000
|
||
00100
|
||
00011
|
||
|
||
Output: 3
|
||
```
|
||
|
||
可以将矩阵表示看成一张有向图。
|
||
|
||
```java
|
||
private int m, n;
|
||
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
|
||
|
||
public int numIslands(char[][] grid) {
|
||
if (grid == null || grid.length == 0) {
|
||
return 0;
|
||
}
|
||
m = grid.length;
|
||
n = grid[0].length;
|
||
int islandsNum = 0;
|
||
for (int i = 0; i < m; i++) {
|
||
for (int j = 0; j < n; j++) {
|
||
if (grid[i][j] != '0') {
|
||
dfs(grid, i, j);
|
||
islandsNum++;
|
||
}
|
||
}
|
||
}
|
||
return islandsNum;
|
||
}
|
||
|
||
private void dfs(char[][] grid, int i, int j) {
|
||
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') {
|
||
return;
|
||
}
|
||
grid[i][j] = '0';
|
||
for (int[] d : direction) {
|
||
dfs(grid, i + d[0], j + d[1]);
|
||
}
|
||
}
|
||
```
|
||
|
||
## 3. 好友关系的连通分量数目
|
||
|
||
[547. Friend Circles (Medium)](https://leetcode.com/problems/friend-circles/description/)
|
||
|
||
```html
|
||
Input:
|
||
[[1,1,0],
|
||
[1,1,0],
|
||
[0,0,1]]
|
||
|
||
Output: 2
|
||
|
||
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
|
||
The 2nd student himself is in a friend circle. So return 2.
|
||
```
|
||
|
||
题目描述：好友关系可以看成是一个无向图，例如第 0 个人与第 1 个人是好友，那么 M[0][1] 和 M[1][0] 的值都为 1。
|
||
|
||
```java
|
||
private int n;
|
||
|
||
public int findCircleNum(int[][] M) {
|
||
n = M.length;
|
||
int circleNum = 0;
|
||
boolean[] hasVisited = new boolean[n];
|
||
for (int i = 0; i < n; i++) {
|
||
if (!hasVisited[i]) {
|
||
dfs(M, i, hasVisited);
|
||
circleNum++;
|
||
}
|
||
}
|
||
return circleNum;
|
||
}
|
||
|
||
private void dfs(int[][] M, int i, boolean[] hasVisited) {
|
||
hasVisited[i] = true;
|
||
for (int k = 0; k < n; k++) {
|
||
if (M[i][k] == 1 && !hasVisited[k]) {
|
||
dfs(M, k, hasVisited);
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
## 4. 填充封闭区域
|
||
|
||
[130. Surrounded Regions (Medium)](https://leetcode.com/problems/surrounded-regions/description/)
|
||
|
||
```html
|
||
For example,
|
||
X X X X
|
||
X O O X
|
||
X X O X
|
||
X O X X
|
||
|
||
After running your function, the board should be:
|
||
X X X X
|
||
X X X X
|
||
X X X X
|
||
X O X X
|
||
```
|
||
|
||
题目描述：使被 'X' 包围的 'O' 转换为 'X'。
|
||
|
||
先填充最外侧，剩下的就是里侧了。
|
||
|
||
```java
|
||
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
|
||
private int m, n;
|
||
|
||
public void solve(char[][] board) {
|
||
if (board == null || board.length == 0) {
|
||
return;
|
||
}
|
||
|
||
m = board.length;
|
||
n = board[0].length;
|
||
|
||
for (int i = 0; i < m; i++) {
|
||
dfs(board, i, 0);
|
||
dfs(board, i, n - 1);
|
||
}
|
||
for (int i = 0; i < n; i++) {
|
||
dfs(board, 0, i);
|
||
dfs(board, m - 1, i);
|
||
}
|
||
|
||
for (int i = 0; i < m; i++) {
|
||
for (int j = 0; j < n; j++) {
|
||
if (board[i][j] == 'T') {
|
||
board[i][j] = 'O';
|
||
} else if (board[i][j] == 'O') {
|
||
board[i][j] = 'X';
|
||
}
|
||
}
|
||
}
|
||
}
|
||
|
||
private void dfs(char[][] board, int r, int c) {
|
||
if (r < 0 || r >= m || c < 0 || c >= n || board[r][c] != 'O') {
|
||
return;
|
||
}
|
||
board[r][c] = 'T';
|
||
for (int[] d : direction) {
|
||
dfs(board, r + d[0], c + d[1]);
|
||
}
|
||
}
|
||
```
|
||
|
||
## 5. 能到达的太平洋和大西洋的区域
|
||
|
||
[417. Pacific Atlantic Water Flow (Medium)](https://leetcode.com/problems/pacific-atlantic-water-flow/description/)
|
||
|
||
```html
|
||
Given the following 5x5 matrix:
|
||
|
||
Pacific ~ ~ ~ ~ ~
|
||
~ 1 2 2 3 (5) *
|
||
~ 3 2 3 (4) (4) *
|
||
~ 2 4 (5) 3 1 *
|
||
~ (6) (7) 1 4 5 *
|
||
~ (5) 1 1 2 4 *
|
||
* * * * * Atlantic
|
||
|
||
Return:
|
||
[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
|
||
```
|
||
|
||
左边和上边是太平洋，右边和下边是大西洋，内部的数字代表海拔，海拔高的地方的水能够流到低的地方，求解水能够流到太平洋和大西洋的所有位置。
|
||
|
||
```java
|
||
|
||
private int m, n;
|
||
private int[][] matrix;
|
||
private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
|
||
|
||
public List<int[]> pacificAtlantic(int[][] matrix) {
|
||
List<int[]> ret = new ArrayList<>();
|
||
if (matrix == null || matrix.length == 0) {
|
||
return ret;
|
||
}
|
||
|
||
m = matrix.length;
|
||
n = matrix[0].length;
|
||
this.matrix = matrix;
|
||
boolean[][] canReachP = new boolean[m][n];
|
||
boolean[][] canReachA = new boolean[m][n];
|
||
|
||
for (int i = 0; i < m; i++) {
|
||
dfs(i, 0, canReachP);
|
||
dfs(i, n - 1, canReachA);
|
||
}
|
||
for (int i = 0; i < n; i++) {
|
||
dfs(0, i, canReachP);
|
||
dfs(m - 1, i, canReachA);
|
||
}
|
||
|
||
for (int i = 0; i < m; i++) {
|
||
for (int j = 0; j < n; j++) {
|
||
if (canReachP[i][j] && canReachA[i][j]) {
|
||
ret.add(new int[]{i, j});
|
||
}
|
||
}
|
||
}
|
||
|
||
return ret;
|
||
}
|
||
|
||
private void dfs(int r, int c, boolean[][] canReach) {
|
||
if (canReach[r][c]) {
|
||
return;
|
||
}
|
||
canReach[r][c] = true;
|
||
for (int[] d : direction) {
|
||
int nextR = d[0] + r;
|
||
int nextC = d[1] + c;
|
||
if (nextR < 0 || nextR >= m || nextC < 0 || nextC >= n
|
||
|| matrix[r][c] > matrix[nextR][nextC]) {
|
||
|
||
continue;
|
||
}
|
||
dfs(nextR, nextC, canReach);
|
||
}
|
||
}
|
||
```
|
||
|
||
# Backtracking
|
||
|
||
Backtracking（回溯）属于 DFS。
|
||
|
||
- 普通 DFS 主要用在 **可达性问题** ，这种问题只需要执行到特点的位置然后返回即可。
|
||
- 而 Backtracking 主要用于求解 **排列组合** 问题，例如有 { 'a','b','c' } 三个字符，求解所有由这三个字符排列得到的字符串，这种问题在执行到特定的位置返回之后还会继续执行求解过程。
|
||
|
||
因为 Backtracking 不是立即返回，而要继续求解，因此在程序实现时，需要注意对元素的标记问题：
|
||
|
||
- 在访问一个新元素进入新的递归调用时，需要将新元素标记为已经访问，这样才能在继续递归调用时不用重复访问该元素；
|
||
- 但是在递归返回时，需要将元素标记为未访问，因为只需要保证在一个递归链中不同时访问一个元素，可以访问已经访问过但是不在当前递归链中的元素。
|
||
|
||
## 1. 数字键盘组合
|
||
|
||
[17. Letter Combinations of a Phone Number (Medium)](https://leetcode.com/problems/letter-combinations-of-a-phone-number/description/)
|
||
|
||
<div align="center"> <img src="pics/9823768c-212b-4b1a-b69a-b3f59e07b977.jpg"/> </div><br>
|
||
|
||
```html
|
||
Input:Digit string "23"
|
||
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
|
||
```
|
||
|
||
```java
|
||
private static final String[] KEYS = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
|
||
|
||
public List<String> letterCombinations(String digits) {
|
||
List<String> combinations = new ArrayList<>();
|
||
if (digits == null || digits.length() == 0) {
|
||
return combinations;
|
||
}
|
||
doCombination(new StringBuilder(), combinations, digits);
|
||
return combinations;
|
||
}
|
||
|
||
private void doCombination(StringBuilder prefix, List<String> combinations, final String digits) {
|
||
if (prefix.length() == digits.length()) {
|
||
combinations.add(prefix.toString());
|
||
return;
|
||
}
|
||
int curDigits = digits.charAt(prefix.length()) - '0';
|
||
String letters = KEYS[curDigits];
|
||
for (char c : letters.toCharArray()) {
|
||
prefix.append(c); // 添加
|
||
doCombination(prefix, combinations, digits);
|
||
prefix.deleteCharAt(prefix.length() - 1); // 删除
|
||
}
|
||
}
|
||
```
|
||
|
||
## 2. IP 地址划分
|
||
|
||
[93. Restore IP Addresses(Medium)](https://leetcode.com/problems/restore-ip-addresses/description/)
|
||
|
||
```html
|
||
Given "25525511135",
|
||
return ["255.255.11.135", "255.255.111.35"].
|
||
```
|
||
|
||
```java
|
||
public List<String> restoreIpAddresses(String s) {
|
||
List<String> addresses = new ArrayList<>();
|
||
StringBuilder tempAddress = new StringBuilder();
|
||
doRestore(0, tempAddress, addresses, s);
|
||
return addresses;
|
||
}
|
||
|
||
private void doRestore(int k, StringBuilder tempAddress, List<String> addresses, String s) {
|
||
if (k == 4 || s.length() == 0) {
|
||
if (k == 4 && s.length() == 0) {
|
||
addresses.add(tempAddress.toString());
|
||
}
|
||
return;
|
||
}
|
||
for (int i = 0; i < s.length() && i <= 2; i++) {
|
||
if (i != 0 && s.charAt(0) == '0') {
|
||
break;
|
||
}
|
||
String part = s.substring(0, i + 1);
|
||
if (Integer.valueOf(part) <= 255) {
|
||
if (tempAddress.length() != 0) {
|
||
part = "." + part;
|
||
}
|
||
tempAddress.append(part);
|
||
doRestore(k + 1, tempAddress, addresses, s.substring(i + 1));
|
||
tempAddress.delete(tempAddress.length() - part.length(), tempAddress.length());
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
## 3. 在矩阵中寻找字符串
|
||
|
||
[79. Word Search (Medium)](https://leetcode.com/problems/word-search/description/)
|
||
|
||
```html
|
||
For example,
|
||
Given board =
|
||
[
|
||
['A','B','C','E'],
|
||
['S','F','C','S'],
|
||
['A','D','E','E']
|
||
]
|
||
word = "ABCCED", -> returns true,
|
||
word = "SEE", -> returns true,
|
||
word = "ABCB", -> returns false.
|
||
```
|
||
|
||
```java
|
||
private final static int[][] direction = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
|
||
private int m;
|
||
private int n;
|
||
|
||
public boolean exist(char[][] board, String word) {
|
||
if (word == null || word.length() == 0) {
|
||
return true;
|
||
}
|
||
if (board == null || board.length == 0 || board[0].length == 0) {
|
||
return false;
|
||
}
|
||
|
||
m = board.length;
|
||
n = board[0].length;
|
||
boolean[][] hasVisited = new boolean[m][n];
|
||
|
||
for (int r = 0; r < m; r++) {
|
||
for (int c = 0; c < n; c++) {
|
||
if (backtracking(0, r, c, hasVisited, board, word)) {
|
||
return true;
|
||
}
|
||
}
|
||
}
|
||
|
||
return false;
|
||
}
|
||
|
||
private boolean backtracking(int curLen, int r, int c, boolean[][] visited, final char[][] board, final String word) {
|
||
if (curLen == word.length()) {
|
||
return true;
|
||
}
|
||
if (r < 0 || r >= m || c < 0 || c >= n
|
||
|| board[r][c] != word.charAt(curLen) || visited[r][c]) {
|
||
|
||
return false;
|
||
}
|
||
|
||
visited[r][c] = true;
|
||
|
||
for (int[] d : direction) {
|
||
if (backtracking(curLen + 1, r + d[0], c + d[1], visited, board, word)) {
|
||
return true;
|
||
}
|
||
}
|
||
|
||
visited[r][c] = false;
|
||
|
||
return false;
|
||
}
|
||
```
|
||
|
||
## 4. 输出二叉树中所有从根到叶子的路径
|
||
|
||
[257. Binary Tree Paths (Easy)](https://leetcode.com/problems/binary-tree-paths/description/)
|
||
|
||
```html
|
||
1
|
||
/ \
|
||
2 3
|
||
\
|
||
5
|
||
```
|
||
|
||
```html
|
||
["1->2->5", "1->3"]
|
||
```
|
||
|
||
```java
|
||
|
||
public List<String> binaryTreePaths(TreeNode root) {
|
||
List<String> paths = new ArrayList<>();
|
||
if (root == null) {
|
||
return paths;
|
||
}
|
||
List<Integer> values = new ArrayList<>();
|
||
backtracking(root, values, paths);
|
||
return paths;
|
||
}
|
||
|
||
private void backtracking(TreeNode node, List<Integer> values, List<String> paths) {
|
||
if (node == null) {
|
||
return;
|
||
}
|
||
values.add(node.val);
|
||
if (isLeaf(node)) {
|
||
paths.add(buildPath(values));
|
||
} else {
|
||
backtracking(node.left, values, paths);
|
||
backtracking(node.right, values, paths);
|
||
}
|
||
values.remove(values.size() - 1);
|
||
}
|
||
|
||
private boolean isLeaf(TreeNode node) {
|
||
return node.left == null && node.right == null;
|
||
}
|
||
|
||
private String buildPath(List<Integer> values) {
|
||
StringBuilder str = new StringBuilder();
|
||
for (int i = 0; i < values.size(); i++) {
|
||
str.append(values.get(i));
|
||
if (i != values.size() - 1) {
|
||
str.append("->");
|
||
}
|
||
}
|
||
return str.toString();
|
||
}
|
||
```
|
||
|
||
## 5. 排列
|
||
|
||
[46. Permutations (Medium)](https://leetcode.com/problems/permutations/description/)
|
||
|
||
```html
|
||
[1,2,3] have the following permutations:
|
||
[
|
||
[1,2,3],
|
||
[1,3,2],
|
||
[2,1,3],
|
||
[2,3,1],
|
||
[3,1,2],
|
||
[3,2,1]
|
||
]
|
||
```
|
||
|
||
```java
|
||
public List<List<Integer>> permute(int[] nums) {
|
||
List<List<Integer>> permutes = new ArrayList<>();
|
||
List<Integer> permuteList = new ArrayList<>();
|
||
boolean[] hasVisited = new boolean[nums.length];
|
||
backtracking(permuteList, permutes, hasVisited, nums);
|
||
return permutes;
|
||
}
|
||
|
||
private void backtracking(List<Integer> permuteList, List<List<Integer>> permutes, boolean[] visited, final int[] nums) {
|
||
if (permuteList.size() == nums.length) {
|
||
permutes.add(new ArrayList<>(permuteList)); // 重新构造一个 List
|
||
return;
|
||
}
|
||
for (int i = 0; i < visited.length; i++) {
|
||
if (visited[i]) {
|
||
continue;
|
||
}
|
||
visited[i] = true;
|
||
permuteList.add(nums[i]);
|
||
backtracking(permuteList, permutes, visited, nums);
|
||
permuteList.remove(permuteList.size() - 1);
|
||
visited[i] = false;
|
||
}
|
||
}
|
||
```
|
||
|
||
## 6. 含有相同元素求排列
|
||
|
||
[47. Permutations II (Medium)](https://leetcode.com/problems/permutations-ii/description/)
|
||
|
||
```html
|
||
[1,1,2] have the following unique permutations:
|
||
[[1,1,2], [1,2,1], [2,1,1]]
|
||
```
|
||
|
||
数组元素可能含有相同的元素，进行排列时就有可能出现重复的排列，要求重复的排列只返回一个。
|
||
|
||
在实现上，和 Permutations 不同的是要先排序，然后在添加一个元素时，判断这个元素是否等于前一个元素，如果等于，并且前一个元素还未访问，那么就跳过这个元素。
|
||
|
||
```java
|
||
public List<List<Integer>> permuteUnique(int[] nums) {
|
||
List<List<Integer>> permutes = new ArrayList<>();
|
||
List<Integer> permuteList = new ArrayList<>();
|
||
Arrays.sort(nums); // 排序
|
||
boolean[] hasVisited = new boolean[nums.length];
|
||
backtracking(permuteList, permutes, hasVisited, nums);
|
||
return permutes;
|
||
}
|
||
|
||
private void backtracking(List<Integer> permuteList, List<List<Integer>> permutes, boolean[] visited, final int[] nums) {
|
||
if (permuteList.size() == nums.length) {
|
||
permutes.add(new ArrayList<>(permuteList));
|
||
return;
|
||
}
|
||
|
||
for (int i = 0; i < visited.length; i++) {
|
||
if (i != 0 && nums[i] == nums[i - 1] && !visited[i - 1]) {
|
||
continue; // 防止重复
|
||
}
|
||
if (visited[i]){
|
||
continue;
|
||
}
|
||
visited[i] = true;
|
||
permuteList.add(nums[i]);
|
||
backtracking(permuteList, permutes, visited, nums);
|
||
permuteList.remove(permuteList.size() - 1);
|
||
visited[i] = false;
|
||
}
|
||
}
|
||
```
|
||
|
||
## 7. 组合
|
||
|
||
[77. Combinations (Medium)](https://leetcode.com/problems/combinations/description/)
|
||
|
||
```html
|
||
If n = 4 and k = 2, a solution is:
|
||
[
|
||
[2,4],
|
||
[3,4],
|
||
[2,3],
|
||
[1,2],
|
||
[1,3],
|
||
[1,4],
|
||
]
|
||
```
|
||
|
||
```java
|
||
public List<List<Integer>> combine(int n, int k) {
|
||
List<List<Integer>> combinations = new ArrayList<>();
|
||
List<Integer> combineList = new ArrayList<>();
|
||
backtracking(combineList, combinations, 1, k, n);
|
||
return combinations;
|
||
}
|
||
|
||
private void backtracking(List<Integer> combineList, List<List<Integer>> combinations, int start, int k, final int n) {
|
||
if (k == 0) {
|
||
combinations.add(new ArrayList<>(combineList));
|
||
return;
|
||
}
|
||
for (int i = start; i <= n - k + 1; i++) { // 剪枝
|
||
combineList.add(i);
|
||
backtracking(combineList, combinations, i + 1, k - 1, n);
|
||
combineList.remove(combineList.size() - 1);
|
||
}
|
||
}
|
||
```
|
||
|
||
## 8. 组合求和
|
||
|
||
[39. Combination Sum (Medium)](https://leetcode.com/problems/combination-sum/description/)
|
||
|
||
```html
|
||
given candidate set [2, 3, 6, 7] and target 7,
|
||
A solution set is:
|
||
[[7],[2, 2, 3]]
|
||
```
|
||
|
||
```java
|
||
public List<List<Integer>> combinationSum(int[] candidates, int target) {
|
||
List<List<Integer>> combinations = new ArrayList<>();
|
||
backtracking(new ArrayList<>(), combinations, 0, target, candidates);
|
||
return combinations;
|
||
}
|
||
|
||
private void backtracking(List<Integer> tempCombination, List<List<Integer>> combinations,
|
||
int start, int target, final int[] candidates) {
|
||
|
||
if (target == 0) {
|
||
combinations.add(new ArrayList<>(tempCombination));
|
||
return;
|
||
}
|
||
for (int i = start; i < candidates.length; i++) {
|
||
if (candidates[i] <= target) {
|
||
tempCombination.add(candidates[i]);
|
||
backtracking(tempCombination, combinations, i, target - candidates[i], candidates);
|
||
tempCombination.remove(tempCombination.size() - 1);
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
## 9. 含有相同元素的组合求和
|
||
|
||
[40. Combination Sum II (Medium)](https://leetcode.com/problems/combination-sum-ii/description/)
|
||
|
||
```html
|
||
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
|
||
A solution set is:
|
||
[
|
||
[1, 7],
|
||
[1, 2, 5],
|
||
[2, 6],
|
||
[1, 1, 6]
|
||
]
|
||
```
|
||
|
||
```java
|
||
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
|
||
List<List<Integer>> combinations = new ArrayList<>();
|
||
Arrays.sort(candidates);
|
||
backtracking(new ArrayList<>(), combinations, new boolean[candidates.length], 0, target, candidates);
|
||
return combinations;
|
||
}
|
||
|
||
private void backtracking(List<Integer> tempCombination, List<List<Integer>> combinations,
|
||
boolean[] hasVisited, int start, int target, final int[] candidates) {
|
||
|
||
if (target == 0) {
|
||
combinations.add(new ArrayList<>(tempCombination));
|
||
return;
|
||
}
|
||
for (int i = start; i < candidates.length; i++) {
|
||
if (i != 0 && candidates[i] == candidates[i - 1] && !hasVisited[i - 1]) {
|
||
continue;
|
||
}
|
||
if (candidates[i] <= target) {
|
||
tempCombination.add(candidates[i]);
|
||
hasVisited[i] = true;
|
||
backtracking(tempCombination, combinations, hasVisited, i + 1, target - candidates[i], candidates);
|
||
hasVisited[i] = false;
|
||
tempCombination.remove(tempCombination.size() - 1);
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
## 10. 1-9 数字的组合求和
|
||
|
||
[216. Combination Sum III (Medium)](https://leetcode.com/problems/combination-sum-iii/description/)
|
||
|
||
```html
|
||
Input: k = 3, n = 9
|
||
|
||
Output:
|
||
|
||
[[1,2,6], [1,3,5], [2,3,4]]
|
||
```
|
||
|
||
从 1-9 数字中选出 k 个数不重复的数，使得它们的和为 n。
|
||
|
||
```java
|
||
public List<List<Integer>> combinationSum3(int k, int n) {
|
||
List<List<Integer>> combinations = new ArrayList<>();
|
||
List<Integer> path = new ArrayList<>();
|
||
backtracking(k, n, 1, path, combinations);
|
||
return combinations;
|
||
}
|
||
|
||
private void backtracking(int k, int n, int start,
|
||
List<Integer> tempCombination, List<List<Integer>> combinations) {
|
||
|
||
if (k == 0 && n == 0) {
|
||
combinations.add(new ArrayList<>(tempCombination));
|
||
return;
|
||
}
|
||
if (k == 0 || n == 0) {
|
||
return;
|
||
}
|
||
for (int i = start; i <= 9; i++) {
|
||
tempCombination.add(i);
|
||
backtracking(k - 1, n - i, i + 1, tempCombination, combinations);
|
||
tempCombination.remove(tempCombination.size() - 1);
|
||
}
|
||
}
|
||
```
|
||
|
||
## 11. 子集
|
||
|
||
[78. Subsets (Medium)](https://leetcode.com/problems/subsets/description/)
|
||
|
||
找出集合的所有子集，子集不能重复，[1, 2] 和 [2, 1] 这种子集算重复
|
||
|
||
```java
|
||
public List<List<Integer>> subsets(int[] nums) {
|
||
List<List<Integer>> subsets = new ArrayList<>();
|
||
List<Integer> tempSubset = new ArrayList<>();
|
||
for (int size = 0; size <= nums.length; size++) {
|
||
backtracking(0, tempSubset, subsets, size, nums); // 不同的子集大小
|
||
}
|
||
return subsets;
|
||
}
|
||
|
||
private void backtracking(int start, List<Integer> tempSubset, List<List<Integer>> subsets,
|
||
final int size, final int[] nums) {
|
||
|
||
if (tempSubset.size() == size) {
|
||
subsets.add(new ArrayList<>(tempSubset));
|
||
return;
|
||
}
|
||
for (int i = start; i < nums.length; i++) {
|
||
tempSubset.add(nums[i]);
|
||
backtracking(i + 1, tempSubset, subsets, size, nums);
|
||
tempSubset.remove(tempSubset.size() - 1);
|
||
}
|
||
}
|
||
```
|
||
|
||
## 12. 含有相同元素求子集
|
||
|
||
[90. Subsets II (Medium)](https://leetcode.com/problems/subsets-ii/description/)
|
||
|
||
```html
|
||
For example,
|
||
If nums = [1,2,2], a solution is:
|
||
|
||
[
|
||
[2],
|
||
[1],
|
||
[1,2,2],
|
||
[2,2],
|
||
[1,2],
|
||
[]
|
||
]
|
||
```
|
||
|
||
```java
|
||
public List<List<Integer>> subsetsWithDup(int[] nums) {
|
||
Arrays.sort(nums);
|
||
List<List<Integer>> subsets = new ArrayList<>();
|
||
List<Integer> tempSubset = new ArrayList<>();
|
||
boolean[] hasVisited = new boolean[nums.length];
|
||
for (int size = 0; size <= nums.length; size++) {
|
||
backtracking(0, tempSubset, subsets, hasVisited, size, nums); // 不同的子集大小
|
||
}
|
||
return subsets;
|
||
}
|
||
|
||
private void backtracking(int start, List<Integer> tempSubset, List<List<Integer>> subsets, boolean[] hasVisited,
|
||
final int size, final int[] nums) {
|
||
|
||
if (tempSubset.size() == size) {
|
||
subsets.add(new ArrayList<>(tempSubset));
|
||
return;
|
||
}
|
||
for (int i = start; i < nums.length; i++) {
|
||
if (i != 0 && nums[i] == nums[i - 1] && !hasVisited[i - 1]) {
|
||
continue;
|
||
}
|
||
tempSubset.add(nums[i]);
|
||
hasVisited[i] = true;
|
||
backtracking(i + 1, tempSubset, subsets, hasVisited, size, nums);
|
||
hasVisited[i] = false;
|
||
tempSubset.remove(tempSubset.size() - 1);
|
||
}
|
||
}
|
||
```
|
||
|
||
## 13. 分割字符串使得每个部分都是回文数
|
||
|
||
[131. Palindrome Partitioning (Medium)](https://leetcode.com/problems/palindrome-partitioning/description/)
|
||
|
||
```html
|
||
For example, given s = "aab",
|
||
Return
|
||
|
||
[
|
||
["aa","b"],
|
||
["a","a","b"]
|
||
]
|
||
```
|
||
|
||
```java
|
||
public List<List<String>> partition(String s) {
|
||
List<List<String>> partitions = new ArrayList<>();
|
||
List<String> tempPartition = new ArrayList<>();
|
||
doPartition(s, partitions, tempPartition);
|
||
return partitions;
|
||
}
|
||
|
||
private void doPartition(String s, List<List<String>> partitions, List<String> tempPartition) {
|
||
if (s.length() == 0) {
|
||
partitions.add(new ArrayList<>(tempPartition));
|
||
return;
|
||
}
|
||
for (int i = 0; i < s.length(); i++) {
|
||
if (isPalindrome(s, 0, i)) {
|
||
tempPartition.add(s.substring(0, i + 1));
|
||
doPartition(s.substring(i + 1), partitions, tempPartition);
|
||
tempPartition.remove(tempPartition.size() - 1);
|
||
}
|
||
}
|
||
}
|
||
|
||
private boolean isPalindrome(String s, int begin, int end) {
|
||
while (begin < end) {
|
||
if (s.charAt(begin++) != s.charAt(end--)) {
|
||
return false;
|
||
}
|
||
}
|
||
return true;
|
||
}
|
||
```
|
||
|
||
## 14. 数独
|
||
|
||
[37. Sudoku Solver (Hard)](https://leetcode.com/problems/sudoku-solver/description/)
|
||
|
||
<div align="center"> <img src="pics/0e8fdc96-83c1-4798-9abe-45fc91d70b9d.png"/> </div><br>
|
||
|
||
```java
|
||
private boolean[][] rowsUsed = new boolean[9][10];
|
||
private boolean[][] colsUsed = new boolean[9][10];
|
||
private boolean[][] cubesUsed = new boolean[9][10];
|
||
private char[][] board;
|
||
|
||
public void solveSudoku(char[][] board) {
|
||
this.board = board;
|
||
for (int i = 0; i < 9; i++)
|
||
for (int j = 0; j < 9; j++) {
|
||
if (board[i][j] == '.') {
|
||
continue;
|
||
}
|
||
int num = board[i][j] - '0';
|
||
rowsUsed[i][num] = true;
|
||
colsUsed[j][num] = true;
|
||
cubesUsed[cubeNum(i, j)][num] = true;
|
||
}
|
||
backtracking(0, 0);
|
||
}
|
||
|
||
private boolean backtracking(int row, int col) {
|
||
while (row < 9 && board[row][col] != '.') {
|
||
row = col == 8 ? row + 1 : row;
|
||
col = col == 8 ? 0 : col + 1;
|
||
}
|
||
if (row == 9) {
|
||
return true;
|
||
}
|
||
for (int num = 1; num <= 9; num++) {
|
||
if (rowsUsed[row][num] || colsUsed[col][num] || cubesUsed[cubeNum(row, col)][num]) {
|
||
continue;
|
||
}
|
||
rowsUsed[row][num] = colsUsed[col][num] = cubesUsed[cubeNum(row, col)][num] = true;
|
||
board[row][col] = (char) (num + '0');
|
||
if (backtracking(row, col)) {
|
||
return true;
|
||
}
|
||
board[row][col] = '.';
|
||
rowsUsed[row][num] = colsUsed[col][num] = cubesUsed[cubeNum(row, col)][num] = false;
|
||
}
|
||
return false;
|
||
}
|
||
|
||
private int cubeNum(int i, int j) {
|
||
int r = i / 3;
|
||
int c = j / 3;
|
||
return r * 3 + c;
|
||
}
|
||
```
|
||
|
||
## 15. N 皇后
|
||
|
||
[51. N-Queens (Hard)](https://leetcode.com/problems/n-queens/description/)
|
||
|
||
<div align="center"> <img src="pics/067b310c-6877-40fe-9dcf-10654e737485.jpg"/> </div><br>
|
||
|
||
在 n\*n 的矩阵中摆放 n 个皇后，并且每个皇后不能在同一行，同一列，同一对角线上，求所有的 n 皇后的解。
|
||
|
||
一行一行地摆放，在确定一行中的那个皇后应该摆在哪一列时，需要用三个标记数组来确定某一列是否合法，这三个标记数组分别为：列标记数组、45 度对角线标记数组和 135 度对角线标记数组。
|
||
|
||
45 度对角线标记数组的长度为 2 \* n - 1，通过下图可以明确 (r, c) 的位置所在的数组下标为 r + c。
|
||
|
||
<div align="center"> <img src="pics/9c422923-1447-4a3b-a4e1-97e663738187.jpg" width="300px"> </div><br>
|
||
|
||
|
||
135 度对角线标记数组的长度也是 2 \* n - 1，(r, c) 的位置所在的数组下标为 n - 1 - (r - c)。
|
||
|
||
<div align="center"> <img src="pics/7a85e285-e152-4116-b6dc-3fab27ba9437.jpg" width="300px"> </div><br>
|
||
|
||
```java
|
||
private List<List<String>> solutions;
|
||
private char[][] nQueens;
|
||
private boolean[] colUsed;
|
||
private boolean[] diagonals45Used;
|
||
private boolean[] diagonals135Used;
|
||
private int n;
|
||
|
||
public List<List<String>> solveNQueens(int n) {
|
||
solutions = new ArrayList<>();
|
||
nQueens = new char[n][n];
|
||
for (int i = 0; i < n; i++) {
|
||
Arrays.fill(nQueens[i], '.');
|
||
}
|
||
colUsed = new boolean[n];
|
||
diagonals45Used = new boolean[2 * n - 1];
|
||
diagonals135Used = new boolean[2 * n - 1];
|
||
this.n = n;
|
||
backtracking(0);
|
||
return solutions;
|
||
}
|
||
|
||
private void backtracking(int row) {
|
||
if (row == n) {
|
||
List<String> list = new ArrayList<>();
|
||
for (char[] chars : nQueens) {
|
||
list.add(new String(chars));
|
||
}
|
||
solutions.add(list);
|
||
return;
|
||
}
|
||
|
||
for (int col = 0; col < n; col++) {
|
||
int diagonals45Idx = row + col;
|
||
int diagonals135Idx = n - 1 - (row - col);
|
||
if (colUsed[col] || diagonals45Used[diagonals45Idx] || diagonals135Used[diagonals135Idx]) {
|
||
continue;
|
||
}
|
||
nQueens[row][col] = 'Q';
|
||
colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = true;
|
||
backtracking(row + 1);
|
||
colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = false;
|
||
nQueens[row][col] = '.';
|
||
}
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
# 微信公众号
|
||
|
||
|
||
更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
|
||
|
||
|
||
<br><div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/公众号海报6.png"></img></div>
|