diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md index d37eba0a..e96fa6f9 100644 --- a/notes/Leetcode 题解.md +++ b/notes/Leetcode 题解.md @@ -1237,9 +1237,9 @@ private int getShortestPath(List[] graphic, int start, int end) {

-广度优先搜索一层一层遍历，每一层得到的所有新节点，要用队列先存储起来以备下一层遍历的时候再遍历。 +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到一个新节点时立马对新节点进行遍历：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。 +而深度优先搜索在得到一个新节点时立马对新节点进行遍历：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。 @@ -1268,29 +1268,29 @@ 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) { + 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++) { + 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) { + 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) { + for (int[] d : direction) area += dfs(grid, r + d[0], c + d[1]); - } + return area; } ``` @@ -1300,11 +1300,13 @@ private int dfs(int[][] grid, int r, int c) { [200. Number of Islands (Medium)](https://leetcode.com/problems/number-of-islands/description/) ```html -11110 -11010 +Input: 11000 -00000 -Answer: 1 +11000 +00100 +00011 + +Output: 3 ``` 可以将矩阵表示看成一张有向图。 @@ -1314,31 +1316,29 @@ 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) { + 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++) { + 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') { + if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') return; - } + grid[i][j] = '0'; - for (int[] d : direction) { + for (int[] d : direction) dfs(grid, i + d[0], j + d[1]); - } } ``` @@ -1365,23 +1365,20 @@ public int findCircleNum(int[][] M) { n = M.length; int circleNum = 0; boolean[] hasVisited = new boolean[n]; - for (int i = 0; i < n; i++) { + 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]) { + for (int k = 0; k < n; k++) + if (M[i][k] == 1 && !hasVisited[k]) dfs(M, k, hasVisited); - } - } } ``` @@ -1403,7 +1400,7 @@ X X X X X O X X ``` -题目描述：使得被 'X' 的 'O' 转换为 'X'。 +题目描述：使得被 'X' 包围的 'O' 转换为 'X'。 先填充最外侧，剩下的就是里侧了。 @@ -1412,9 +1409,12 @@ 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; + 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); @@ -1423,24 +1423,27 @@ public void solve(char[][] board) { dfs(board, 0, i); dfs(board, m - 1, i); } - for (int i = 0; i < m; 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'; + 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; + if (r < 0 || r >= m || c < 0 || c >= n || board[r][c] != 'O') + return; + board[r][c] = 'T'; - for (int[] d : direction) { + for (int[] d : direction) dfs(board, r + d[0], c + d[1]); - } } ``` -**从两个方向都能到达的区域** +**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/) @@ -1468,12 +1471,15 @@ private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; public List pacificAtlantic(int[][] matrix) { List ret = new ArrayList<>(); - if (matrix == null || matrix.length == 0) return ret; + 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); @@ -1482,24 +1488,25 @@ public List pacificAtlantic(int[][] matrix) { 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]) { + + 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; + 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; + if (nextR < 0 || nextR >= m || nextC < 0 || nextC >= n || matrix[r][c] > matrix[nextR][nextC]) + continue; dfs(nextR, nextC, canReach); } } @@ -1533,7 +1540,8 @@ private static final String[] KEYS = {"", "", "abc", "def", "ghi", "jkl", "mno", public List letterCombinations(String digits) { List ret = new ArrayList<>(); - if (digits == null || digits.length() == 0) return ret; + if (digits == null || digits.length() == 0) + return ret; combination(new StringBuilder(), digits, ret); return ret; } @@ -1571,16 +1579,17 @@ public List restoreIpAddresses(String s) { private void doRestore(int k, StringBuilder path, String s, List addresses) { if (k == 4 || s.length() == 0) { - if (k == 4 && s.length() == 0) { + if (k == 4 && s.length() == 0) addresses.add(path.toString()); - } return; } for (int i = 0; i < s.length() && i <= 2; i++) { - if (i != 0 && s.charAt(0) == '0') break; + if (i != 0 && s.charAt(0) == '0') + break; String part = s.substring(0, i + 1); if (Integer.valueOf(part) <= 255) { - if (path.length() != 0) part = "." + part; + if (path.length() != 0) + part = "." + part; path.append(part); doRestore(k + 1, path, s.substring(i + 1), addresses); path.delete(path.length() - part.length(), path.length()); @@ -1612,33 +1621,36 @@ 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; + 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[][] visited = new boolean[m][n]; - for (int i = 0; i < m; i++) { - for (int j = 0; j < n; j++) { + + for (int i = 0; i < m; i++) + for (int j = 0; j < n; j++) if (backtracking(board, visited, word, 0, i, j)) return true; - } - } + return false; } private boolean backtracking(char[][] board, boolean[][] visited, String word, int start, int r, int c) { - if (start == word.length()) { + if (start == word.length()) return true; - } - if (r < 0 || r >= m || c < 0 || c >= n || board[r][c] != word.charAt(start) || visited[r][c]) { + if (r < 0 || r >= m || c < 0 || c >= n || board[r][c] != word.charAt(start) || visited[r][c]) return false; - } + visited[r][c] = true; - for (int[] d : direction) { - if (backtracking(board, visited, word, start + 1, r + d[0], c + d[1])) { + + for (int[] d : direction) + if (backtracking(board, visited, word, start + 1, r + d[0], c + d[1])) return true; - } - } + visited[r][c] = false; + return false; } ``` @@ -1662,18 +1674,20 @@ private boolean backtracking(char[][] board, boolean[][] visited, String word, i ```java public List binaryTreePaths(TreeNode root) { List paths = new ArrayList(); - if (root == null) return paths; + if (root == null) + return paths; List values = new ArrayList<>(); backtracking(root, values, paths); return paths; } private void backtracking(TreeNode node, List values, List paths) { - if (node == null) return; + if (node == null) + return; values.add(node.val); - if (isLeaf(node)) { + if (isLeaf(node)) paths.add(buildPath(values)); - } else { + else { backtracking(node.left, values, paths); backtracking(node.right, values, paths); } @@ -1688,9 +1702,8 @@ private String buildPath(List values) { StringBuilder str = new StringBuilder(); for (int i = 0; i < values.size(); i++) { str.append(values.get(i)); - if (i != values.size() - 1) { + if (i != values.size() - 1) str.append("->"); - } } return str.toString(); } @@ -1727,7 +1740,8 @@ private void backtracking(List permuteList, boolean[] visited, int[] nu return; } for (int i = 0; i < visited.length; i++) { - if (visited[i]) continue; + if (visited[i]) + continue; visited[i] = true; permuteList.add(nums[i]); backtracking(permuteList, visited, nums, ret); @@ -1767,8 +1781,10 @@ private void backtracking(List permuteList, boolean[] visited, int[] nu } 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; + 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, visited, nums, ret); @@ -1827,27 +1843,25 @@ A solution set is: ``` ```java - private List> ret; +public List> combinationSum(int[] candidates, int target) { + List> ret = new ArrayList<>(); + doCombination(candidates, target, 0, new ArrayList<>(), ret); + return ret; +} - public List> combinationSum(int[] candidates, int target) { - ret = new ArrayList<>(); - doCombination(candidates, target, 0, new ArrayList<>()); - return ret; - } - - private void doCombination(int[] candidates, int target, int start, List list) { - if (target == 0) { - ret.add(new ArrayList<>(list)); - return; - } - for (int i = start; i < candidates.length; i++) { - if (candidates[i] <= target) { - list.add(candidates[i]); - doCombination(candidates, target - candidates[i], i, list); - list.remove(list.size() - 1); - } - } - } +private void doCombination(int[] candidates, int target, int start, List list, List> ret) { + if (target == 0) { + ret.add(new ArrayList<>(list)); + return; + } + for (int i = start; i < candidates.length; i++) { + if (candidates[i] <= target) { + list.add(candidates[i]); + doCombination(candidates, target - candidates[i], i, list, ret); + list.remove(list.size() - 1); + } + } +} ``` **含有相同元素的求组合求和** @@ -1866,26 +1880,25 @@ A solution set is: ``` ```java -private List> ret; - public List> combinationSum2(int[] candidates, int target) { - ret = new ArrayList<>(); + List> ret = new ArrayList<>(); Arrays.sort(candidates); - doCombination(candidates, target, 0, new ArrayList<>(), new boolean[candidates.length]); + doCombination(candidates, target, 0, new ArrayList<>(), new boolean[candidates.length], ret); return ret; } -private void doCombination(int[] candidates, int target, int start, List list, boolean[] visited) { +private void doCombination(int[] candidates, int target, int start, List list, boolean[] visited, List> ret) { if (target == 0) { ret.add(new ArrayList<>(list)); return; } for (int i = start; i < candidates.length; i++) { - if (i != 0 && candidates[i] == candidates[i - 1] && !visited[i - 1]) continue; + if (i != 0 && candidates[i] == candidates[i - 1] && !visited[i - 1]) + continue; if (candidates[i] <= target) { list.add(candidates[i]); visited[i] = true; - doCombination(candidates, target - candidates[i], i + 1, list, visited); + doCombination(candidates, target - candidates[i], i + 1, list, visited, ret); visited[i] = false; list.remove(list.size() - 1); } @@ -1905,17 +1918,13 @@ Output: [[1,2,6], [1,3,5], [2,3,4]] ``` -题目描述：从 1-9 数字中选出 k 个数，使得它们的和为 n。 +题目描述：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> combinationSum3(int k, int n) { List> ret = new ArrayList<>(); List path = new ArrayList<>(); - for (int i = 1; i <= 9; i++) { - path.add(i); - backtracking(k - 1, n - i, path, i, ret); - path.remove(0); - } + backtracking(k, n, path, 1, ret); return ret; } @@ -1924,10 +1933,11 @@ private void backtracking(int k, int n, List path, int start, List(path)); return; } - if (k == 0 || n == 0) return; - for (int i = start + 1; i <= 9; i++) { // 只能访问下一个元素，防止遍历的结果重复 + if (k == 0 || n == 0) + return; + for (int i = start; i <= 9; i++) { path.add(i); - backtracking(k - 1, n - i, path, i, ret); + backtracking(k - 1, n - i, path, i + 1, ret); path.remove(path.size() - 1); } } @@ -1946,9 +1956,8 @@ private List subsetList; public List> subsets(int[] nums) { ret = new ArrayList<>(); subsetList = new ArrayList<>(); - for (int i = 0; i <= nums.length; i++) { // 不同的子集大小 + for (int i = 0; i <= nums.length; i++) // 不同的子集大小 backtracking(0, i, nums); - } return ret; } @@ -1957,7 +1966,6 @@ private void backtracking(int startIdx, int size, int[] nums) { ret.add(new ArrayList(subsetList)); return; } - for (int i = startIdx; i < nums.length; i++) { subsetList.add(nums[i]); backtracking(i + 1, size, nums); @@ -1994,9 +2002,10 @@ public List> subsetsWithDup(int[] nums) { subsetList = new ArrayList<>(); visited = new boolean[nums.length]; Arrays.sort(nums); - for (int i = 0; i <= nums.length; i++) { + + for (int i = 0; i <= nums.length; i++) backtracking(0, i, nums); - } + return ret; } @@ -2005,9 +2014,9 @@ private void backtracking(int startIdx, int size, int[] nums) { ret.add(new ArrayList(subsetList)); return; } - for (int i = startIdx; i < nums.length; i++) { - if (i != 0 && nums[i] == nums[i - 1] && !visited[i - 1]) continue; + if (i != 0 && nums[i] == nums[i - 1] && !visited[i - 1]) + continue; subsetList.add(nums[i]); visited[i] = true; backtracking(i + 1, size, nums); @@ -2036,11 +2045,11 @@ private List> ret; public List> partition(String s) { ret = new ArrayList<>(); - doPartion(new ArrayList<>(), s); + doPartition(new ArrayList<>(), s); return ret; } -private void doPartion(List list, String s) { +private void doPartition(List list, String s) { if (s.length() == 0) { ret.add(new ArrayList<>(list)); return; @@ -2048,16 +2057,16 @@ private void doPartion(List list, String s) { for (int i = 0; i < s.length(); i++) { if (isPalindrome(s, 0, i)) { list.add(s.substring(0, i + 1)); - doPartion(list, s.substring(i + 1)); + doPartition(list, s.substring(i + 1)); list.remove(list.size() - 1); } } } private boolean isPalindrome(String s, int begin, int end) { - while (begin < end) { - if (s.charAt(begin++) != s.charAt(end--)) return false; - } + while (begin < end) + if (s.charAt(begin++) != s.charAt(end--)) + return false; return true; } ``` @@ -2076,20 +2085,19 @@ private char[][] board; public void solveSudoku(char[][] board) { this.board = board; - for (int i = 0; i < 9; i++) { + for (int i = 0; i < 9; i++) for (int j = 0; j < 9; j++) { - if (board[i][j] == '.') continue; + 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; } - } - for (int i = 0; i < 9; i++) { - for (int j = 0; j < 9; j++) { + + for (int i = 0; i < 9; i++) + for (int j = 0; j < 9; j++) backtracking(i, j); - } - } } private boolean backtracking(int row, int col) { @@ -2097,14 +2105,17 @@ private boolean backtracking(int row, int col) { row = col == 8 ? row + 1 : row; col = col == 8 ? 0 : col + 1; } - if (row == 9) { + + 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; + 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; + if (backtracking(row, col)) + return true; board[row][col] = '.'; rowsUsed[row][num] = colsUsed[col][num] = cubesUsed[cubeNum(row, col)][num] = false; } @@ -2124,15 +2135,15 @@ private int cubeNum(int i, int j) {

-题目描述：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，要求解所有的 n 皇后解。 +题目描述：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维度为 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。 +45 度对角线标记数组的维度为 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。

-135 度对角线标记数组的维度也是 2\*n - 1，(r,c) 的位置所在的数组下标为 n - 1 - (r - c)。 +135 度对角线标记数组的维度也是 2 \* n - 1，(r, c) 的位置所在的数组下标为 n - 1 - (r - c)。

@@ -2147,21 +2158,21 @@ private int n; public List> solveNQueens(int n) { ret = new ArrayList<>(); nQueens = new char[n][n]; - Arrays.fill(nQueens, '.'); + 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; - backstracking(0); + backtracking(0); return ret; } -private void backstracking(int row) { +private void backtracking(int row) { if (row == n) { List list = new ArrayList<>(); - for (char[] chars : nQueens) { + for (char[] chars : nQueens) list.add(new String(chars)); - } ret.add(list); return; } @@ -2169,12 +2180,11 @@ private void backstracking(int row) { for (int col = 0; col < n; col++) { int diagonals45Idx = row + col; int diagonals135Idx = n - 1 - (row - col); - if (colUsed[col] || diagonals45Used[diagonals45Idx] || diagonals135Used[diagonals135Idx]) { + if (colUsed[col] || diagonals45Used[diagonals45Idx] || diagonals135Used[diagonals135Idx]) continue; - } nQueens[row][col] = 'Q'; colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = true; - backstracking(row + 1); + backtracking(row + 1); colUsed[col] = diagonals45Used[diagonals45Idx] = diagonals135Used[diagonals135Idx] = false; nQueens[row][col] = '.'; } diff --git a/notes/设计模式.md b/notes/设计模式.md index 93cdd0da..400ede5e 100644 --- a/notes/设计模式.md +++ b/notes/设计模式.md @@ -233,7 +233,7 @@ public class Client { ```java public abstract class Factory { abstract public Product factoryMethod(); - public void doSomethind() { + public void doSomething() { Product product = factoryMethod(); // do something with the product }