代码随想录训练营第23天|回溯去重

39. 组合总和

class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    int sum=0;

    void dfs(vector<int>& candidates, int target, int startIdx){
        if(sum==target){
            result.push_back(path);
            return;
        }
        if(sum>target)
            return;
        for(int i=startIdx; i<candidates.size(); i++){
            path.push_back(candidates[i]);
            sum+=candidates[i];
            dfs(candidates,target,i);
            sum-=candidates[i];
            path.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        dfs(candidates,target,0);
        return result;
    }
};

因为可以重复选取，所以递归的下一个数字的搜索起点（startIdx）是i，而非i+1。

40. 组合总和 II

class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    int sum=0;

    void dfs(vector<int>& candidates, int target, int startIdx){
        if(sum==target){
            result.push_back(path);
            return;
        }
        if(sum>target)
            return;
        for(int i=startIdx; i< candidates.size();i++){
            path.push_back(candidates[i]);
            sum+=candidates[i];
            dfs(candidates,target,i+1);
            sum-=candidates[i];
            path.pop_back();
            while(i+1<candidates.size()&&candidates[i+1]==candidates[i])
                i++;
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        dfs(candidates,target,0);
        return result;
    }
};

回溯之后，相同数值的元素直接跳过，实现本层去重功能。 

 131. 分割回文串

class Solution {
public:
    vector<vector<string>> result;
    vector<string> path;

    bool check(string& s) {
        int left = 0, right = s.length() - 1;
        while (left < right) {
            if (s[left++] != s[right--])
                return false;
        }
        return true;
    }

    void dfs(string& s, int stardIdx) {
        if (stardIdx == s.length()) {
            result.push_back(path);
            return;
        }
        for (int endIdx = stardIdx + 1; endIdx <= s.length(); endIdx++) {
            string sub = s.substr(stardIdx, endIdx - stardIdx); //[startIdx,endIdx) 左闭右开
            if (check(sub)) {
                path.push_back(sub);
                dfs(s, endIdx);
                path.pop_back();
            }
        }
    }
    vector<vector<string>> partition(string s) {
        dfs(s, 0);
        return result;
    }
};

子串在合法的情况下，才会尝试递归新的起点位置（startIdx)，直到已经切分到末尾，收集结果。