时间：2021-10-28 10:25:39 | 栏目：C代码 | 点击：次
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:
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]
]
这道题跟之前那道 Combination Sum 本质没有区别，只需要改动一点点即可，之前那道题给定数组中的数字可以重复使用，而这道题不能重复使用，只需要在之前的基础上修改两个地方即可，首先在递归的 for 循环里加上 if (i > start && num[i] == num[i - 1]) continue; 这样可以防止 res 中出现重复项，然后就在递归调用 helper 里面的参数换成 i+1，这样就不会重复使用数组中的数字了，代码如下：
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& num, int target) {
vector<vector<int>> res;
vector<int> out;
sort(num.begin(), num.end());
helper(num, target, 0, out, res);
return res;
}
void helper(vector<int>& num, int target, int start, vector<int>& out, vector<vector<int>>& res) {
if (target < 0) return;
if (target == 0) { res.push_back(out); return; }
for (int i = start; i < num.size(); ++i) {
if (i > start && num[i] == num[i - 1]) continue;
out.push_back(num[i]);
helper(num, target - num[i], i + 1, out, res);
out.pop_back();
}
}
};