算法day20-回溯（2）

目录

1. 组合总和
2. 组合总和II
3. 分割回文串 ⭐⭐⭐

一、组合总和

 https://leetcode.cn/problems/combination-sum/description/?envType=problem-list-v2&envId=8At1GmaZ

class Solution {
    List<List<Integer>> res;
    List<Integer> path;
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        res = new ArrayList<>();
        path = new ArrayList<>();
        dfs(0, target, candidates, 0);
        return res;
    }

    public void dfs(int i, int target, int[] candidates,int sum){
        if(sum == target){           //终止条件
            res.add(new ArrayList<>(path));
            return;
        }
        if(sum > target){
            return;
        }
        
        for(int j=i; j<candidates.length; j++){
            sum += candidates[j];
            path.add(candidates[j]);
            dfs(j, target, candidates,sum);
            path.remove(path.size()-1);
            sum -= candidates[j];
        }
    }
}

 

二、组合总和II

 https://leetcode.cn/problems/combination-sum-ii/?envType=problem-list-v2&envId=8At1GmaZ

 

class Solution {
    List<List<Integer>> res;
    List<Integer> path;
    int[] visited;
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        res = new ArrayList<>();
        path = new ArrayList<>();
        visited = new int[candidates.length];
        Arrays.sort(candidates);
        dfs(0, candidates, target, 0);
        return res;
    }
    public void dfs(int i, int[] candidates, int target, int sum){
        if(sum == target){
            res.add(new ArrayList<>(path));
            return;
        }
        if(sum > target){
            return;
        }
        for(int j=i; j<candidates.length; j++){
            // 对于每一层，如果有重复的数字-->跳过选择
            // 假如第一层选定了，第二层选过了x，那么这个第二层就不能再选x了
            if(j > i && candidates[j] == candidates[j - 1]) continue;
            path.add(candidates[j]);
            dfs(j+1, candidates, target, sum + candidates[j]);
            path.remove(path.size()-1);
        }
    }
}

三、分割回文串

 https://leetcode.cn/problems/palindrome-partitioning/?envType=problem-list-v2&envId=8At1GmaZ

class Solution {
    List<List<String>> res;
    List<String> path;

    public List<List<String>> partition(String s) {
        res = new ArrayList<>();
        path = new ArrayList<>();
        dfs(0, s);
        return res;
    }

    public void dfs(int start, String s){
        if(start == s.length()){
            res.add(new ArrayList<>(path));
            return;
        }
        for(int end = start; end<s.length();end++){
            if (isPalindrome(s, start, end)) {
                path.add(s.substring(start, end+1));
                dfs(end + 1, s);
                path.remove(path.size()-1);
            }
        }
    }

    public boolean isPalindrome(String s, int start, int end){
        if(s == null){
            return true;
        }
        while(start < end){
            if(s.charAt(start) != s.charAt(end)){
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
}