博客记录-day146-力扣（回溯）

一、力扣

1、反转字符串中的单词

151. 反转字符串中的单词

class Solution {
    public String reverseWords(String s) {
        int left = 0, right = s.length() - 1;
        // 去掉字符串开头的空白字符
        while (left <= right && s.charAt(left) == ' ') {
            ++left;
        }

        // 去掉字符串末尾的空白字符
        while (left <= right && s.charAt(right) == ' ') {
            --right;
        }

        Deque<String> d = new ArrayDeque<String>();
        StringBuilder word = new StringBuilder();
        
        while (left <= right) {
            char c = s.charAt(left);
            if ((word.length() != 0) && (c == ' ')) {
                // 将单词 push 到队列的头部
                d.offerFirst(word.toString());
                word.setLength(0);
            } else if (c != ' ') {
                word.append(c);
            }
            ++left;
        }
        d.offerFirst(word.toString());

        return String.join(" ", d);
    }
}

2、最小栈

155. 最小栈

class MinStack {
    ArrayDeque<int[]> stack;

    public MinStack() {
        stack=new ArrayDeque<>();
        stack.push(new int[]{0, Integer.MAX_VALUE});
    }
    
    public void push(int val) {
        int[] ans=stack.peek();
        stack.push(new int[]{val,Math.min(ans[1],val)});
    }
    
    public void pop() {
        stack.pop();
    }
    
    public int top() {
        return stack.peek()[0];
    }
    
    public int getMin() {
        return stack.peek()[1];
    }
}

3、组合总和

39. 组合总和

class Solution {
    List<List<Integer>> res;
    List<Integer> path;
    int[] candidates;
    int target;
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        res=new ArrayList<>();
        path=new ArrayList<>();
        this.candidates=candidates;
        this.target=target;
        dfs(0,0);
        return res;
    }
    public void dfs(int start,int sum){
        if(sum>target) return;
        if(sum==target){
            res.add(new ArrayList<>(path));
            return;
        }
        for(int i=start;i<candidates.length;i++){
            path.add(candidates[i]);
            dfs(i,sum+candidates[i]);
            path.removeLast();
        }
    }
}

4、组合总和 II

40. 组合总和 II

本题的难点在于：集合（数组candidates）有重复元素，但还不能有重复的组合。

class Solution {
    List<List<Integer>> res;
    List<Integer> path;
    int[] candidates;
    int target;

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        res = new ArrayList<>();
        path = new ArrayList<>();
        Arrays.sort(candidates);
        int[] vis = new int[candidates.length];
        this.candidates = candidates;
        this.target = target;
        dfs(0, 0, vis);
        return res;
    }

    public void dfs(int start, int sum, int[] vis) {
        if (sum > target)
            return;
        if (sum == target) {
            res.add(new ArrayList<>(path));
            return;
        }
        for (int i = start; i < candidates.length; i++) {
            if (i > 0 && candidates[i] == candidates[i - 1] && vis[i - 1] == 0)
                continue;
            path.add(candidates[i]);
            vis[i] = 1;
            dfs(i + 1, sum + candidates[i], vis);
            path.removeLast();
            vis[i] = 0;
        }
    }
}

5、组合总和 III

216. 组合总和 III

class Solution {
    List<List<Integer>> res;
    List<Integer> path;
    int k;
    int n;

    public List<List<Integer>> combinationSum3(int k, int n) {
        res = new ArrayList<>();
        path = new ArrayList<>();
        this.k = k;
        this.n = n;
        dfs(1, 0);
        return res;
    }

    public void dfs(int start, int sum) {
        if (sum > n || path.size() > k)
            return;
        if (sum == n && path.size() == k) {
            res.add(new ArrayList<>(path));
            return;
        }
        for (int i = start; i <= 9; i++) {
            path.add(i);
            dfs(i + 1, sum + i);
            path.removeLast();
        }
    }
}

6、电话号码的字母组合

17. 电话号码的字母组合

class Solution {
    List<String> result=new ArrayList<>();
    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) {
            return result;
        }
        //初始对应所有的数字，为了直接对应2-9，新增了两个无效的字符串""
        String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        //迭代处理
        backTracking(digits, numString, 0);
        return result;
    }
    StringBuilder temp = new StringBuilder();
    //比如digits如果为"23",num 为0，则str表示2对应的 abc
    public void backTracking(String digits, String[] numString, int num) {
        //遍历全部一次记录一次得到的字符串
        if (num == digits.length()) {
            result.add(temp.toString());
            return;
        }
        //str 表示当前num对应的字符串
        String str = numString[digits.charAt(num) - '0'];
        for (int i = 0; i < str.length(); i++) {
            temp.append(str.charAt(i));
            //递归，处理下一层
            backTracking(digits, numString, num + 1);
            //剔除末尾的继续尝试
            temp.deleteCharAt(temp.length() - 1);
        }
    }
}