LeetCode 热题 HOT 100：回溯专题

LeetCode 热题 HOT 100：https://leetcode.cn/problem-list/2cktkvj/

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17. 电话号码的字母组合

题目链接：https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {
    List<String> list;
    Map<Character, String> map;

    public List<String> letterCombinations(String digits) {
        ap = new HashMap<>();
        res = new LinkedList<>();
        if(digits.length() == 0){
            return res;
        }
        map.put('2', "abc");
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");
        backTrack(digits, 0, new StringBuilder());
        return list;
    }

    public void backTrack(String digits, int ind, StringBuilder sb){ // 参数：输入的字符串、字符串的索引、拼接的英文字符串
        if(digits.length() == ind){
            list.add(sb.toString());
        }else{
            char ch = digits.charAt(ind);
            String str = map.get(ch); // 获取按键下面的字符序列
            for (int i = 0; i < str.length(); i++) {
                sb.append(str.charAt(i));
                backTrack(digits, ind + 1, sb);
                sb.deleteCharAt(sb.length()-1); // 回溯
            }
        }
    }
}

参考题解：https://leetcode.cn/problems/letter-combinations-of-a-phone-number/solutions/388738/dian-hua-hao-ma-de-zi-mu-zu-he-by-leetcode-solutio/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

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22. 括号生成

题目链接：https://leetcode.cn/problems/generate-parentheses/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

思路一：

class Solution {
    List<String> res;
    public List<String> generateParenthesis(int n) {
        res = new LinkedList<>();
        backTrack(n, 0, 0, "");
        return res;
    }

    public void backTrack(int n, int left, int right, String str){
        if(left < right){
            return;
        }
        if(left == n && right == n){
            res.add(str);
            return;
        }else{
            if(left < n){
                backTrack(n, left+1, right, str+"(");
            }
            backTrack(n, left, right+1, str+")");
        }
    }
}

思路二：

class Solution {
    List<String> res = new ArrayList<>();
    public List<String> generateParenthesis(int n) {
        if(n<=0){
            return res;
        }
        getBracket("", n, n);
        return res;
    }

    public void getBracket(String str, int left, int right){
        if(left == 0 && right == 0){
            res.add(str);
            return;
        }
        if(left == right){
            getBracket(str+"(", left-1, right);
        }else{
            if(left > 0){
                getBracket(str+"(", left-1, right);
            }
            getBracket(str+")", left, right-1);
        }
    }
}

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39. 组合总和

题目链接：https://leetcode.cn/problems/combination-sum/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> list = new ArrayList<>();

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        dfs(candidates, target, 0);
        return res;
    }

    public void dfs(int[] candidates, int target, int ind){ // 关键点在于索引
        if(target == 0){
            res.add(new ArrayList<>(list));
            return;
        }

        for(int i = ind; i < candidates.length; i ++){
            if(target - candidates[i] >= 0){
                list.add(candidates[i]);
                dfs(candidates, target - candidates[i], i); 
			    // 每次在当前的索引上进行遍历，作用在于：如果没有索引：target=5，5-3-2 作用等同于 5-2-3, 那么会有两种组合[2,3]与[3,2]
                // 但是在索引的约束下，不会出现这种情况 
                list.remove(list.size()-1);
            }
        }
    }
}

参考链接：https://leetcode.cn/problems/combination-sum/solutions/14697/hui-su-suan-fa-jian-zhi-python-dai-ma-java-dai-m-2/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

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46. 全排列

题目链接：https://leetcode.cn/problems/permutations/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> list = new ArrayList<>();

    public List<List<Integer>> permute(int[] nums) {
        boolean[] visited = new boolean[nums.length]; // 标志数组
        dfs(nums, 0, visited);
        return res;
    }

    public void dfs(int[] nums, int size, boolean[] visited){
        if(size == nums.length){
            res.add(new ArrayList<>(list));
            return;
        }
        for(int i = 0; i < nums.length; i ++){
            if(!visited[i]){
                visited[i] = true;
                list.add(nums[i]);
                dfs(nums, size+1, visited);
                list.remove(list.size()-1); // 回溯
                visited[i] = false;
            }
        }
    }
}

参考链接：https://leetcode.cn/problems/permutations/solutions/9914/hui-su-suan-fa-python-dai-ma-java-dai-ma-by-liweiw/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

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补充：47. 全排列 II （待优化)

题目链接：https://leetcode.cn/problems/permutations-ii/description/?envType=featured-list&envId=2cktkvj%3FenvType%3Dfeatured-list&envId=2cktkvj

class Solution {
    List<List<Integer>> res = new LinkedList<>();
    List<Integer> list = new LinkedList<>();

    public List<List<Integer>> permuteUnique(int[] nums) {
        boolean[] visited = new boolean[nums.length]; // 标志数组
        dfs(nums, 0, visited);
        return res;
    }

    public void dfs(int[] nums, int size, boolean[] visited){
        if(size == nums.length){
            for (List<Integer> result : res) {
                if(result.equals(list)){
                    return;
                }
            }
            res.add(new ArrayList<>(list));
            return;
        }
        for(int i = 0; i < nums.length; i ++){
            if(!visited[i]){
                visited[i] = true;
                list.add(nums[i]);
                dfs(nums, size+1, visited);
                list.remove(list.size()-1);
                visited[i] = false;
            }
        }
    }
}

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78. 子集

题目链接：https://leetcode.cn/problems/subsets/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> list = new ArrayList<>();

    public List<List<Integer>> subsets(int[] nums) {
        dfs(nums, 0);
        return res;
    }

    public void dfs(int[] nums, int i){
        if(i==nums.length){
            res.add(new ArrayList(list));
            return;
        }
        // 选 nums[i]
        list.add(nums[i]);
        dfs(nums, i+1);
        // 不选 nums[i]
        list.remove(list.size()-1);
        dfs(nums, i+1);
    }
}

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79. 单词搜索

题目链接：https://leetcode.cn/problems/word-search/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {

    public boolean exist(char[][] board, String word) {
        for(int i = 0; i < board.length; i ++){
            for(int j = 0; j < board[0].length; j ++){
                if(board[i][j] == word.charAt(0)){
                    if(dfs(board, i, j, word, 0)){ // 路径开头不一定只有一处，所以要遍历整个数组
                        return true;
                    }
                }
            }
        }
        return false;
    }

    public boolean dfs(char[][] board, int i, int j, String word, int ind){
        if(ind >= word.length()){
            return true;
        }
        if(i>=0 && i<board.length && j>=0 && j<board[0].length && board[i][j]==word.charAt(ind) && board[i][j]!='\0'){ // 剪枝
            char tmp = board[i][j];
            board[i][j] = '\0'; // 设置不可访问
            boolean f1 = dfs(board, i, j-1, word, ind+1); // 左
            boolean f2 = dfs(board, i-1, j, word, ind+1); // 上
            boolean f3 = dfs(board, i, j+1, word, ind+1); // 右
            boolean f4 = dfs(board, i+1, j, word, ind+1); // 下
            board[i][j] = tmp; // 回溯
            return f1 || f2 || f3 ||f4;
        }
        return false;
    }
}

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124. 二叉树中的最大路径和

题目链接：https://leetcode.cn/problems/binary-tree-maximum-path-sum/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

• 二叉树 abc，a 是根结点（递归中的 root），bc 是左右子结点（代表其递归后的最优解）。最大的路径，可能的路径情况：
     a
 /  \
  b    c
  ① b + a + c。
  ② b + a + a 的父结点。（需要再次递归）
  ③ a + c + a 的父结点。（需要再次递归）
• 其中情况 1，表示如果不联络父结点的情况，或本身是根结点的情况。这种情况是没法递归的，但是结果有可能是全局最大路径和，因此可以在递归过程中通过比较得出。
• 情况 2 和 3，递归时计算 a+b 和 a+c，选择一个更优的方案返回，也就是上面说的递归后的最优解。

class Solution {
    int max = Integer.MIN_VALUE;

    public int maxPathSum(TreeNode root) {
        if(root == null){
            return 0;
        }
        dfs(root);
        return max;
    }

    /**
     * 返回经过root的单边分支最大和， 即 Math.max(root, root+left, root+right)
     */
    public int dfs(TreeNode root){
        if(root == null){
            return 0;
        }
        // 计算左子树最大值，左边分支如果为负数还不如不选择
        int leftMax = Math.max(0, dfs(root.left));
        // 计算右子树最大值，右边分支如果为负数还不如不选择
        int rightMax = Math.max(0, dfs(root.right));

        // left->root->right 作为路径与已经计算过历史最大值做比较
        max = Math.max(max, leftMax + root.val + rightMax);

        // 返回经过root的单边最大分支给当前root的父节点计算使用
        return root.val + Math.max(leftMax, rightMax);
    }
}

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200. 岛屿数量

题目链接：https://leetcode.cn/problems/number-of-islands/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {
    public int numIslands(char[][] grid) {
        int sum = 0;
        for(int i = 0; i < grid.length; i ++){
            for(int j = 0; j < grid[0].length; j ++){
                if(grid[i][j] == '1'){
                    dfs(grid, i, j);
                    sum++;
                }
            }
        }
        return sum;
    }
    
    public void dfs(char[][] grid, int x, int y){
        if(0<=x && x<grid.length && 0<=y && y<grid[0].length && grid[x][y] == '1'){
            grid[x][y] ='0';
            dfs(grid, x, y-1); // 左
            dfs(grid, x-1, y); // 上
            dfs(grid, x, y+1); // 右
            dfs(grid, x+1, y); // 下
        }
    }
}

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437. 路径总和 III

题目链接：https://leetcode.cn/problems/path-sum-iii/description/?envType=featured-list&envId=2cktkvj?envType=featured-list&envId=2cktkvj

class Solution {
    // key 是前缀和，value 是前缀和为这个值的路径数量。
    Map<Long, Integer> map = new HashMap<>();
    int target;

    public int pathSum(TreeNode root, int targetSum) {
        this.target = targetSum;
        // 可能路径从根节点开始算
        map.put(0l, 1);
        return dfs(root, 0l);
    }

    public int dfs(TreeNode root, long curSum){
        if(root == null){
            return 0;
        }
        curSum += root.val; // 当前累计的结点值
        int res = 0;
        // 以当前节点为止，去查看从前的 map 集合中是否还存在目标前缀和
        //              1
        //             /
        //            2
        //           /
        //          3
        // 假设目标和为 5
        // 节点 1 的前缀和为：1
        // 节点 3 的前缀和为: 1+2+3 = 6
        // pre(3) - pre(1) = 5
        // 所以从节点 1 到节点 3 之间有一条符合要求的路径
        res += map.getOrDefault(curSum-target, 0);

        // 存储路径的原因是可能节点的前缀和存在相等的情况：
        //              2
        //             /
        //            0
        //           /
        //          4
        // 从节点 2 到节点 4 有两条路径长度等于2
        map.put(curSum, map.getOrDefault(curSum, 0) + 1);

        int left = dfs(root.left, curSum); // 调用左子树
        int right = dfs(root.right, curSum); // 调用右子树

        res = res + left + right;

        map.put(curSum, map.get(curSum)-1); // 恢复状态

        return res;
    }
}

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