https://leetcode.com/problems/combination-sum/description/

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

一眼看成了记录方案的背包

递推想了半天不知道咋写

看了topic说是回溯

吭哧吭哧把代码写了

一直RE

把erase(end())换成ans.pop_back();变成了WA

参考别人代码，发现return应在这个句后面对

A了

WA的代码

class Solution {
public:
    vector<vector<int>> tot;
    void dfs(int x,vector<int>& c,int target,vector<int> ans,int s){
        for(int i=s;i<c.size();i++){
            ans.push_back(c[i]);
            if(x+c[i]<target)
                dfs(x+c[i],c,target,ans,i);
            else if(x+c[i]==target){
                tot.push_back(ans);
                return;
            }
            else 
                return ;
            ans.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> ans;
        ans.empty();
        tot.empty();
        dfs(0,candidates,target,ans,0);
        return tot;
    }
};

A的代码

class Solution {
public:
    vector<vector<int>> tot;
    void dfs(int x,vector<int>& c,int target,vector<int> ans,int s){
        for(int i=s;i<c.size();i++){
            ans.push_back(c[i]);
            if(x+c[i]<target)
                dfs(x+c[i],c,target,ans,i);
            else if(x+c[i]==target){
                tot.push_back(ans);
            }
            ans.pop_back();
        }
        return;
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> ans;
        ans.empty();
        tot.empty();
        dfs(0,candidates,target,ans,0);
        return tot;
    }
};