440 lines
12 KiB
Markdown
440 lines
12 KiB
Markdown
<!-- GFM-TOC -->
|
||
* [1. 把数组中的 0 移到末尾](#1-把数组中的-0-移到末尾)
|
||
* [2. 改变矩阵维度](#2-改变矩阵维度)
|
||
* [3. 找出数组中最长的连续 1](#3-找出数组中最长的连续-1)
|
||
* [4. 有序矩阵查找](#4-有序矩阵查找)
|
||
* [5. 有序矩阵的 Kth Element](#5-有序矩阵的-kth-element)
|
||
* [6. 一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数](#6-一个数组元素在-[1,-n]-之间，其中一个数被替换为另一个数，找出重复的数和丢失的数)
|
||
* [7. 找出数组中重复的数，数组值在 [1, n] 之间](#7-找出数组中重复的数，数组值在-[1,-n]-之间)
|
||
* [8. 数组相邻差值的个数](#8-数组相邻差值的个数)
|
||
* [9. 数组的度](#9-数组的度)
|
||
* [10. 对角元素相等的矩阵](#10-对角元素相等的矩阵)
|
||
* [11. 嵌套数组](#11-嵌套数组)
|
||
* [12. 分隔数组](#12-分隔数组)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
# 1. 把数组中的 0 移到末尾
|
||
|
||
[283. Move Zeroes (Easy)](https://leetcode.com/problems/move-zeroes/description/)
|
||
|
||
```html
|
||
For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
|
||
```
|
||
|
||
```java
|
||
public void moveZeroes(int[] nums) {
|
||
int idx = 0;
|
||
for (int num : nums) {
|
||
if (num != 0) {
|
||
nums[idx++] = num;
|
||
}
|
||
}
|
||
while (idx < nums.length) {
|
||
nums[idx++] = 0;
|
||
}
|
||
}
|
||
```
|
||
|
||
# 2. 改变矩阵维度
|
||
|
||
[566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
|
||
|
||
```html
|
||
Input:
|
||
nums =
|
||
[[1,2],
|
||
[3,4]]
|
||
r = 1, c = 4
|
||
|
||
Output:
|
||
[[1,2,3,4]]
|
||
|
||
Explanation:
|
||
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
|
||
```
|
||
|
||
```java
|
||
public int[][] matrixReshape(int[][] nums, int r, int c) {
|
||
int m = nums.length, n = nums[0].length;
|
||
if (m * n != r * c) {
|
||
return nums;
|
||
}
|
||
int[][] reshapedNums = new int[r][c];
|
||
int index = 0;
|
||
for (int i = 0; i < r; i++) {
|
||
for (int j = 0; j < c; j++) {
|
||
reshapedNums[i][j] = nums[index / n][index % n];
|
||
index++;
|
||
}
|
||
}
|
||
return reshapedNums;
|
||
}
|
||
```
|
||
|
||
# 3. 找出数组中最长的连续 1
|
||
|
||
[485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)
|
||
|
||
```java
|
||
public int findMaxConsecutiveOnes(int[] nums) {
|
||
int max = 0, cur = 0;
|
||
for (int x : nums) {
|
||
cur = x == 0 ? 0 : cur + 1;
|
||
max = Math.max(max, cur);
|
||
}
|
||
return max;
|
||
}
|
||
```
|
||
|
||
# 4. 有序矩阵查找
|
||
|
||
[240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
|
||
|
||
```html
|
||
[
|
||
[ 1, 5, 9],
|
||
[10, 11, 13],
|
||
[12, 13, 15]
|
||
]
|
||
```
|
||
|
||
```java
|
||
public boolean searchMatrix(int[][] matrix, int target) {
|
||
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
|
||
int m = matrix.length, n = matrix[0].length;
|
||
int row = 0, col = n - 1;
|
||
while (row < m && col >= 0) {
|
||
if (target == matrix[row][col]) return true;
|
||
else if (target < matrix[row][col]) col--;
|
||
else row++;
|
||
}
|
||
return false;
|
||
}
|
||
```
|
||
|
||
# 5. 有序矩阵的 Kth Element
|
||
|
||
[378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
|
||
|
||
```html
|
||
matrix = [
|
||
[ 1, 5, 9],
|
||
[10, 11, 13],
|
||
[12, 13, 15]
|
||
],
|
||
k = 8,
|
||
|
||
return 13.
|
||
```
|
||
|
||
解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
|
||
|
||
二分查找解法：
|
||
|
||
```java
|
||
public int kthSmallest(int[][] matrix, int k) {
|
||
int m = matrix.length, n = matrix[0].length;
|
||
int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
|
||
while (lo <= hi) {
|
||
int mid = lo + (hi - lo) / 2;
|
||
int cnt = 0;
|
||
for (int i = 0; i < m; i++) {
|
||
for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
|
||
cnt++;
|
||
}
|
||
}
|
||
if (cnt < k) lo = mid + 1;
|
||
else hi = mid - 1;
|
||
}
|
||
return lo;
|
||
}
|
||
```
|
||
|
||
堆解法：
|
||
|
||
```java
|
||
public int kthSmallest(int[][] matrix, int k) {
|
||
int m = matrix.length, n = matrix[0].length;
|
||
PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
|
||
for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
|
||
for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
|
||
Tuple t = pq.poll();
|
||
if(t.x == m - 1) continue;
|
||
pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
|
||
}
|
||
return pq.poll().val;
|
||
}
|
||
|
||
class Tuple implements Comparable<Tuple> {
|
||
int x, y, val;
|
||
public Tuple(int x, int y, int val) {
|
||
this.x = x; this.y = y; this.val = val;
|
||
}
|
||
|
||
@Override
|
||
public int compareTo(Tuple that) {
|
||
return this.val - that.val;
|
||
}
|
||
}
|
||
```
|
||
|
||
# 6. 一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数
|
||
|
||
[645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
|
||
|
||
```html
|
||
Input: nums = [1,2,2,4]
|
||
Output: [2,3]
|
||
```
|
||
|
||
```html
|
||
Input: nums = [1,2,2,4]
|
||
Output: [2,3]
|
||
```
|
||
|
||
最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
|
||
|
||
主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。
|
||
|
||
```java
|
||
public int[] findErrorNums(int[] nums) {
|
||
for (int i = 0; i < nums.length; i++) {
|
||
while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
|
||
swap(nums, i, nums[i] - 1);
|
||
}
|
||
}
|
||
for (int i = 0; i < nums.length; i++) {
|
||
if (nums[i] != i + 1) {
|
||
return new int[]{nums[i], i + 1};
|
||
}
|
||
}
|
||
return null;
|
||
}
|
||
|
||
private void swap(int[] nums, int i, int j) {
|
||
int tmp = nums[i];
|
||
nums[i] = nums[j];
|
||
nums[j] = tmp;
|
||
}
|
||
```
|
||
|
||
类似题目：
|
||
|
||
- [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
|
||
- [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
|
||
|
||
# 7. 找出数组中重复的数，数组值在 [1, n] 之间
|
||
|
||
[287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
|
||
|
||
要求不能修改数组，也不能使用额外的空间。
|
||
|
||
二分查找解法：
|
||
|
||
```java
|
||
public int findDuplicate(int[] nums) {
|
||
int l = 1, h = nums.length - 1;
|
||
while (l <= h) {
|
||
int mid = l + (h - l) / 2;
|
||
int cnt = 0;
|
||
for (int i = 0; i < nums.length; i++) {
|
||
if (nums[i] <= mid) cnt++;
|
||
}
|
||
if (cnt > mid) h = mid - 1;
|
||
else l = mid + 1;
|
||
}
|
||
return l;
|
||
}
|
||
```
|
||
|
||
双指针解法，类似于有环链表中找出环的入口：
|
||
|
||
```java
|
||
public int findDuplicate(int[] nums) {
|
||
int slow = nums[0], fast = nums[nums[0]];
|
||
while (slow != fast) {
|
||
slow = nums[slow];
|
||
fast = nums[nums[fast]];
|
||
}
|
||
fast = 0;
|
||
while (slow != fast) {
|
||
slow = nums[slow];
|
||
fast = nums[fast];
|
||
}
|
||
return slow;
|
||
}
|
||
```
|
||
|
||
# 8. 数组相邻差值的个数
|
||
|
||
[667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
|
||
|
||
```html
|
||
Input: n = 3, k = 2
|
||
Output: [1, 3, 2]
|
||
Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
|
||
```
|
||
|
||
题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
|
||
|
||
让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
|
||
|
||
```java
|
||
public int[] constructArray(int n, int k) {
|
||
int[] ret = new int[n];
|
||
ret[0] = 1;
|
||
for (int i = 1, interval = k; i <= k; i++, interval--) {
|
||
ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
|
||
}
|
||
for (int i = k + 1; i < n; i++) {
|
||
ret[i] = i + 1;
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
# 9. 数组的度
|
||
|
||
[697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
|
||
|
||
```html
|
||
Input: [1,2,2,3,1,4,2]
|
||
Output: 6
|
||
```
|
||
|
||
题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
|
||
|
||
```java
|
||
public int findShortestSubArray(int[] nums) {
|
||
Map<Integer, Integer> numsCnt = new HashMap<>();
|
||
Map<Integer, Integer> numsLastIndex = new HashMap<>();
|
||
Map<Integer, Integer> numsFirstIndex = new HashMap<>();
|
||
for (int i = 0; i < nums.length; i++) {
|
||
int num = nums[i];
|
||
numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
|
||
numsLastIndex.put(num, i);
|
||
if (!numsFirstIndex.containsKey(num)) {
|
||
numsFirstIndex.put(num, i);
|
||
}
|
||
}
|
||
int maxCnt = 0;
|
||
for (int num : nums) {
|
||
maxCnt = Math.max(maxCnt, numsCnt.get(num));
|
||
}
|
||
int ret = nums.length;
|
||
for (int i = 0; i < nums.length; i++) {
|
||
int num = nums[i];
|
||
int cnt = numsCnt.get(num);
|
||
if (cnt != maxCnt) continue;
|
||
ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
# 10. 对角元素相等的矩阵
|
||
|
||
[766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
|
||
|
||
```html
|
||
1234
|
||
5123
|
||
9512
|
||
|
||
In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.
|
||
```
|
||
|
||
```java
|
||
public boolean isToeplitzMatrix(int[][] matrix) {
|
||
for (int i = 0; i < matrix[0].length; i++) {
|
||
if (!check(matrix, matrix[0][i], 0, i)) {
|
||
return false;
|
||
}
|
||
}
|
||
for (int i = 0; i < matrix.length; i++) {
|
||
if (!check(matrix, matrix[i][0], i, 0)) {
|
||
return false;
|
||
}
|
||
}
|
||
return true;
|
||
}
|
||
|
||
private boolean check(int[][] matrix, int expectValue, int row, int col) {
|
||
if (row >= matrix.length || col >= matrix[0].length) {
|
||
return true;
|
||
}
|
||
if (matrix[row][col] != expectValue) {
|
||
return false;
|
||
}
|
||
return check(matrix, expectValue, row + 1, col + 1);
|
||
}
|
||
```
|
||
|
||
# 11. 嵌套数组
|
||
|
||
[565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
|
||
|
||
```html
|
||
Input: A = [5,4,0,3,1,6,2]
|
||
Output: 4
|
||
Explanation:
|
||
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
|
||
|
||
One of the longest S[K]:
|
||
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
|
||
```
|
||
|
||
题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。
|
||
|
||
```java
|
||
public int arrayNesting(int[] nums) {
|
||
int max = 0;
|
||
for (int i = 0; i < nums.length; i++) {
|
||
int cnt = 0;
|
||
for (int j = i; nums[j] != -1; ) {
|
||
cnt++;
|
||
int t = nums[j];
|
||
nums[j] = -1; // 标记该位置已经被访问
|
||
j = t;
|
||
|
||
}
|
||
max = Math.max(max, cnt);
|
||
}
|
||
return max;
|
||
}
|
||
```
|
||
|
||
# 12. 分隔数组
|
||
|
||
[769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems/max-chunks-to-make-sorted/description/)
|
||
|
||
```html
|
||
Input: arr = [1,0,2,3,4]
|
||
Output: 4
|
||
Explanation:
|
||
We can split into two chunks, such as [1, 0], [2, 3, 4].
|
||
However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
|
||
```
|
||
|
||
题目描述：分隔数组，使得对每部分排序后数组就为有序。
|
||
|
||
```java
|
||
public int maxChunksToSorted(int[] arr) {
|
||
if (arr == null) return 0;
|
||
int ret = 0;
|
||
int right = arr[0];
|
||
for (int i = 0; i < arr.length; i++) {
|
||
right = Math.max(right, arr[i]);
|
||
if (right == i) ret++;
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
</br><div align="center">😢 ☹️🙁😐🙂😄😂🤣 </br></br> 更多精彩内容将发布在公众号 **CyC2018**，公众号提供了该项目的离线阅读版本，后台回复"下载" 即可领取。也提供了一份技术面试复习思维导图，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"资料" 即可领取。我基本是按照这个思维导图来进行复习的，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思维导图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。</div></br>
|
||
<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
|