2019.2.15
题目描述：
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]
这题和昨天的https://blog.csdn.net/weixin_41637618/article/details/87250472差不多，就是要求数组中的数不可以重复使用了而已，所以思想上还是差不多的。
解法一：
还是回溯算法，修改一下两个地方，先对原数组进行排序，遇到相同的数字就跳过，这样res中就不会有重复项了，第二个是将递归算法中i+1，就可以保证数字不重复使用了。
C++代码：
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> out;
sort(candidates.begin(),candidates.end());
combinationSum2DFS(candidates,target,0,out,res);
return res;
}
void combinationSum2DFS(vector<int> &candidates,int target,int index,vector<int> &out,vector<vector<int>> &res){
if(target<0) return;
if(target==0){
res.push_back(out);
return;
}
for (int i = index; i < candidates.size(); ++i) {
if(i>index&&candidates[i-1]==candidates[i]) continue;
out.push_back(candidates[i]);
combinationSum2DFS(candidates, target-candidates[i], i+1,out,res);
out.pop_back();
}
}
};