题目链接：
https://leetcode-cn.com/problems/combination-sum/
难度：中等
39. 组合总和
给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的数字可以无限制重复被选取。
说明：
所有数字（包括 target）都是正整数。
解集不能包含重复的组合。
示例 1：
输入：candidates = [2,3,6,7], target = 7,
所求解集为：
[
[7],
[2,2,3]
]
示例 2：
输入：candidates = [2,3,5], target = 8,
所求解集为：
[
[2,2,2,2],
[2,3,3],
[3,5]
]
提示：
1 <= candidates.length <= 30
1 <= candidates[i] <= 200
candidate 中的每个元素都是独一无二的。
1 <= target <= 500
It’s not difficult to backtrack + recursion without considering a lot of questions.
But I made a mistake when I wrote it. Record: Numbers can be reused. There are two cases to consider when recursing: 1. Use the number 2. Do not use this number,
but in the second case, note that pos does not need to be +1 because it can be reused
(ps: extremely low efficiency)
class Solution {
public:
vector<vector<int>> ans;
void dfs(int curtarget,vector<int>& candidates,int pos,vector<int> tmp){
if(curtarget==0){
ans.push_back(tmp);
return;
}
if(pos==candidates.size()||curtarget<0){
return;
}
dfs(curtarget,candidates,pos+1,tmp);
tmp.push_back(candidates[pos]);
dfs(curtarget-candidates[pos],candidates,pos,tmp);
tmp.pop_back();
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<int> tmp;
dfs(target,candidates,0,tmp);
return ans;
}
};
Using pruning is not familiar with pruning,,, the writing is a bit ugly
class Solution {
public:
void dfs(int curtarget, vector<int>& candidates, vector<int> &tmp, int begin, int len,vector<vector<int>>& ans) {
if (curtarget == 0) {
ans.push_back(tmp);
return;
}
for (int i = begin; i < len; ++i) {
int t=candidates[i];
if (curtarget - t < 0) {
break;
}
tmp.push_back(t);
dfs(curtarget - t, candidates, tmp, i, len,ans);
tmp.pop_back();
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<int> tmp;
vector<vector<int>> ans;
sort(candidates.begin(), candidates.end());
dfs(target, candidates, tmp, 0, candidates.size(),ans);
return ans;
}
};