组合问题

LeetCode39: 组合总和

LeetCode39. 组合总和

目标和，除了累加所有的数外还可以用目标值减去所有的数。

添加第i个元素后，可以继续添加第i个元素。可以添加第i个元素，也可以添加索引为candidates.length-1的元素

这类回溯的问题可以想象成多叉数，对于根节点有左右子树，对于组合而言，多叉树的集合是candidates的所有的元素。以及考虑所有子元素的下一层的子元素集合是什么。

class Solution {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> combinationSum(int[] candidates, int target) {backtrack(candidates, target, 0);return res;}private void backtrack(int[] candidates, int target, int idx) {if (target == 0) {res.add(new ArrayList<>(path));}if (target < 0 || idx == candidates.length) {return;}for (int i = idx; i < candidates.length; i++) {path.add(candidates[i]);backtrack(candidates, target - candidates[i], i );path.pollLast();}}
}

LeetCode40: 组合总和II

LeetCode40: 组合总和II

每个数字只能使用一次，并不意味着不允许重复元素。如果原数组中存在两个1，和是2，是可以有[1,1]的组合

进入下一次递归的时候，需要进行索引加1，当前元素不能再用。

i > idx && candidates[i] == candidates[i - 1]，目的是防止数组中有两个1，第一个1和第二个1在与其他元素搭配的时候作用是一样的，需要过滤。因此，数组必须是有序的，保证相等的元素相邻。

class Solution {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> combinationSum2(int[] candidates, int target) {Arrays.sort(candidates);backtrack(candidates, target, 0);return res;}private void backtrack(int[] candidates, int target, int idx) {if (target == 0) {res.add(new ArrayList<>(path));}if (target < 0) {return;}for (int i = idx; i < candidates.length; i++) {if (i > idx && candidates[i] == candidates[i - 1]) {continue;}path.add(candidates[i]);backtrack(candidates, target - candidates[i], i + 1);path.pollLast();}}}

LeetCode17. 电话号码的字母组合

LeetCode17. 电话号码的字母组合

当长度相等满足，则加入结果集

字符’2’转为数字2，digits.charAt(idx) - '0'

class Solution {List<String> res = new ArrayList<>();StringBuilder sb = new StringBuilder();String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};public List<String> letterCombinations(String digits) {if (digits == null || digits.length() == 0) {return res;}dfs(digits, 0);return res;}private void dfs(String digits, int idx) {if (sb.length() == digits.length()) {res.add(sb.toString());return;}String str = numString[digits.charAt(idx) - '0'];for (int i = 0; i < str.length(); i++) {sb.append(str.charAt(i));dfs(digits, idx + 1);sb.deleteCharAt(sb.length() - 1);}}
}

子集问题

LeetCode78：子集

LeetCode78. 子集

startIndex == nums.length这个校验可有可无。

子集第一层子集合是[0,nums.length-1]，对于选择0的元素，下一层的子集合是[1,nums.length-1]

class Solution {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> subsets(int[] nums) {dfs(nums, 0);return res;}private void dfs(int[] nums, int idx) {res.add(new ArrayList<>(path));if (startIndex == nums.length) {return;}for (int i = idx; i < nums.length; i++) {path.add(nums[i]);dfs(nums, i + 1);path.pollLast();}}
}

LeetCode90：子集II

LeetCode90：子集II

i > idx && nums[i] == nums[i - 1]可以理解为同一层元素不能重复。i==idx的时候，i是第一个元素。

需要对数组进行排序，排序好的数组元素相等的在一起。

class Solution {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> subsetsWithDup(int[] nums) {Arrays.sort(nums);dfs(nums, 0);return res;}private void dfs(int[] nums, int idx) {res.add(new ArrayList<>(path));for (int i = idx; i < nums.length; i++) {if (i > idx && nums[i] == nums[i - 1]) {continue;}path.add(nums[i]);dfs(nums, i + 1);path.pollLast();}}
}

LeetCode491: 非递减子序列

LeetCode491: 非递减子序列

子序列就是不能改变数组中的元素。

去重：使用boolean数组或者HashMap

递增的判断是获取path中最后一个元素和当前元素比较。

class Solution {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> findSubsequences(int[] nums) {dfs(nums, 0);return res;}private void dfs(int[] nums, int idx) {if (path.size() > 1) {res.add(new ArrayList<>(path));}if (idx == nums.length) {return;}boolean[] used = new boolean[201];for (int i = idx; i < nums.length; i++) {if (used[nums[i] + 100] || (!path.isEmpty() && path.peekLast() > nums[i])) {continue;}path.add(nums[i]);used[nums[i]+ 100] = true;dfs(nums, i + 1);path.pollLast();}}
}

全排列问题

LeetCode46. 全排列

LeetCode46. 全排列

第一层集合是[0,nums.length-1]，第二层集合是[0,nums.length-1]再减去第一层用的元素。

class Solution {List<List<Integer>> res = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();boolean[] used;public List<List<Integer>> permute(int[] nums) {used = new boolean[nums.length];dfs(nums);return res;}private void dfs(int[] nums) {if (path.size() == nums.length) {res.add(new ArrayList<>(path));return;}for (int i = 0; i < nums.length; i++) {if (used[i]) {continue;}path.add(nums[i]);used[i] = true;dfs(nums);used[i] = false;path.removeLast();}}
}

LeetCode47: 全排列II

LeetCode47: 全排列II

不能有重复的组合

i > 0 && nums[i] == nums[i - 1] && !vis[i - 1]。连续三个1，第一层使用第一个1，第二层使用第二个1（used[i-1]==true）。等于是1(a),1(b),1(c)只能有一个顺序。

class Solution {List<List<Integer>> result = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();boolean[] used;public List<List<Integer>> permuteUnique(int[] nums) {Arrays.sort(nums);used = new boolean[nums.length];dfs(nums);return result;}private void dfs(int[] nums) {if (nums.length == path.size()) {result.add(new ArrayList<>(path));return;}for (int i = 0; i < nums.length; i++) {if (used[i] || (i > 0 && nums[i] == nums[i - 1]) && !used[i - 1]) {continue;}path.add(nums[i]);used[i] = true;dfs(nums);used[i] = false;path.pollLast();}}
}

切割问题

LeetCode93: 复原IP地址（**）

LeetCode93: 复原IP地址

判断某字符串是否为合法字符串，大于等于0，小于等于255的字符串。

要切分为合理的4段，记录start为字符串的起始字符。当符合要求的时候，start == s.length() && split == 4

子树集合，[start,start + 2]，只有3个位置可选。

题目的难点，需要把应用题理解为可以做回溯的题目。

class Solution {List<String> res = new ArrayList<>();LinkedList<String> path = new LinkedList<>();public List<String> restoreIpAddresses(String s) {dfs(s, 0, 0);return res;}private void dfs(String s, int start, int split) {if (start == s.length() && split == 4) {res.add(String.join(".", path));return;}for (int i = start; i < start + 3; i++) {if (i >= s.length()) {break;}if ((4 - split) * 3 < s.length() - i) {continue;}if (isValid(s, start, i)) {path.add(s.substring(start, i + 1));dfs(s, i + 1, split + 1);path.removeLast();}}}private boolean isValid(String s, int start, int end) {if (start != end && s.charAt(start) == '0') {return false;}int res = 0;for (int i = start; i <= end; i++) {res = res * 10 + (s.charAt(i) - '0');}return res >= 0 && res <= 255;}
}

网格类问题

LeetCode200: 岛屿数量

LeetCode200: 岛屿数量

回溯方法：计数，如果当前点为岛屿，将附近的土地置为0，避免重复计数。

class Solution {public int numIslands(char[][] grid) {if (grid == null || grid.length == 0 || grid[0].length == 0) {return 0;}int count = 0;int m = grid.length; //行的个数int n = grid[0].length; // 列的个数for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == '1') {count++;dfs(grid, i, j);}}}return count;}private void dfs(char[][] grid, int i, int j) {if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0') {return;}grid[i][j] = '0';dfs(grid, i - 1, j);dfs(grid, i + 1, j);dfs(grid, i, j - 1);dfs(grid, i, j + 1);}
}

LeetCode695: 岛屿的最大面积

LeetCode695: 岛屿的最大面积

class Solution {public int maxAreaOfIsland(int[][] grid) {if (grid == null || grid.length == 0 || grid[0].length == 0) {return 0;}int m = grid.length;int n = grid[0].length;int ans = 0;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == 1) {ans = Math.max(ans, dfs(grid, i, j));}}}return ans;}private int dfs(int[][] grid, int i, int j) {int res = 1;if (i >= grid.length || i < 0 || j < 0 || j >= grid[0].length || grid[i][j] == 0) {return 0;}grid[i][j] = 0;res += dfs(grid, i - 1, j);res += dfs(grid, i + 1, j);res += dfs(grid, i, j - 1);res += dfs(grid, i, j + 1);return res;}
}

LeetCode79. 单词搜索

LeetCode79. 单词搜索

board[i][j] == '.'设置为该点位已经使用过。

比较索引为index的元素，判断是否相等，相等则继续，不等则返回结果false。比较前后左右的结果。当索引为最后一个元素，判断为相等，则返回结果。

class Solution {public boolean exist(char[][] board, String word) {int m = board.length;int n = board[0].length;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (dfs(board, word, 0, i, j)) {return true;}}}return false;}private boolean dfs(char[][] board, String word, int index, int i, int j) {if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || word.charAt(index) != board[i][j]|| board[i][j] == '.') {return false;}if (index == word.length() - 1) {return true;}char tmp = board[i][j];board[i][j] = '.';boolean b = dfs(board, word, index + 1, i + 1, j)|| dfs(board, word, index + 1, i - 1, j)|| dfs(board, word, index + 1, i, j + 1)|| dfs(board, word, index + 1, i, j - 1);board[i][j] = tmp;return b;}
}

LeetCode207. N皇后

51. N 皇后

定义一个n*n的数组，插入所有的网格 .。

校验新增的元素是否合理（竖向比较，45度比较，横向比较，135度比较）

将当前board保存到List<String>中，加入结果集

class Solution {List<List<String>> res = new ArrayList<>();public List<List<String>> solveNQueens(int n) {char[][] board = new char[n][n];for (int i = 0; i < board.length; i++) {Arrays.fill(board[i], '.');}dfs(board, 0, n);return res;}private void dfs(char[][] board, int index, int n) {if (index == n) {res.add(toArray(board));return;}for (int i = 0; i < board[index].length; i++) {if (isValid(board, index, i, n)) {board[index][i] = 'Q';dfs(board, index + 1, n);board[index][i] = '.';}}}private boolean isValid(char[][] board, int row, int col, int n) {for (int i = 0; i < row; i++) {if (board[i][col] == 'Q') {return false;}}for (int i = 0; i < col; i++) {if (board[row][i] == 'Q') {return false;}}for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {if (board[i][j] == 'Q') {return false;}}for (int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {if (board[i][j] == 'Q') {return false;}}return true;}public List<String> toArray(char[][] board) {List<String> list = new ArrayList<>();for (int i = 0; i < board.length; i++) {list.add(new String(board[i]));}return list;}
}

括号问题

LeetCode22. 括号生成

LeetCode22. 括号生成

这条题目我看到想过并且看答案已经不止3次了，每次看都像看新的题目一样，感觉很熟悉，就是不会做。

记录左括号的次数和右括号的次数，当左括号的次数和右括号的次数都等于n，满足条件。当左括号的次数小于右括号的次数，进行剪枝，比如())

class Solution {List<String> res = new ArrayList<>();public List<String> generateParenthesis(int n) {dfs(0, 0, n, "");return res;}private void dfs(int left, int right, int n, String s) {if (left == n && right == n) {res.add(s);return;}if (left < right) {return;}if (left < n) {dfs(left + 1, right, n, s + "(");}if (right < n) {dfs(left, right + 1, n, s + ")");}}
}

图问题

LeetCode207. 课程表

LeetCode207. 课程表

记录每个课程的先学课程数量，没有先学课程的加入队列。

记录边，边是[先学,后学]

class Solution {public boolean canFinish(int numCourses, int[][] prerequisites) {int[] indegree = new int[numCourses];List<List<Integer>> edges = new ArrayList<>();for (int i = 0; i < numCourses; i++) {edges.add(new ArrayList<>());}for (int[] prereq : prerequisites) {indegree[prereq[0]]++; // 入度,[1,2] 先学2再学1edges.get(prereq[1]).add(prereq[0]);}Queue<Integer> queue = new LinkedList<>();for (int i = 0; i < indegree.length; i++) {if (indegree[i] == 0) {queue.add(i);}}while (!queue.isEmpty()) {int poll = queue.poll();numCourses--;for (Integer pre : edges.get(poll)) {indegree[pre]--;if (indegree[pre] == 0) {queue.add(pre);}}}return numCourses == 0;}
}