[代码随想录23回溯]回溯的组合问题＋分割子串

前言

回溯是什么？

题目链接

39. 组合总和 - 力扣（LeetCode）

40. 组合总和 II - 力扣（LeetCode）

131. 分割回文串 - 力扣（LeetCode）

一、组合问题

    private:
    vector<vector<int>>res;
    vector<int> path;
    void backtracking(vector<int>&candidates,int target,int sum,int startIndex){
        if(sum==target){
            res.push_back(path);
            return ;
        }
        for(int i=startIndex;i<candidates.size()&&sum+candidates[i]<=target;i++){
            sum+=candidates[i];
            path.push_back(candidates[i]);
            backtracking(candidates,target,sum,i);
            sum-=candidates[i];
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        res.clear();
        path.clear();
        sort(candidates.begin(),candidates.end());
        backtracking(candidates,target,0,0);
        return res;
    }

二、组合问题

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

三、分割回文串

    vector<vector<string>> res;
    vector<string>path;
    void backtracking(const string&s,int startIndex){
        if(startIndex>=s.size()){
            res.push_back(path);
            return ;
        }
        for(int i=startIndex;i<s.size();i++){
            if(isPalindrome(s,startIndex,i)){
                string str=s.substr(startIndex,i-startIndex+1);
                path.push_back(str);
            }else{
                continue;
            }
            backtracking(s,i+1);
            path.pop_back();
        }
    }
    bool isPalindrome(const string& s,int start,int end){
        for(int i=start,j=end;i<j;i++,j--){
            if(s[i]!=s[j]) return false;
        }
        return true;
    }

public:
    vector<vector<string>> partition(string s) {
        backtracking(s,0);
        return res;
    }

总结