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]
]题目链接：https://leetcode-cn.com/problems/combination-sum-ii/
思路
法一：回溯法
相比第39题，多了2个条件：会有重复元素，每个元素最多用一次。
去重：需要对重复元素产生的重复结果去重，在同一轮迭代的循环中，遇到重复的元素直接跳过。
剪枝：因为初始时对数组进行了递增排序，下一层迭代只需要从下个元素开始即可。
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
if(candidates.size()<=0) return res;
sort(candidates.begin(), candidates.end());
vector<int> cur;
trace(candidates, target, cur, -1);
return res;
}
void trace(vector<int> candidates, int target, vector<int> &cur, int idx){
for(int i=idx+1; i<candidates.size(); ++i){
if(candidates[i]>target) break;
if((i>idx+1 && candidates[i]==candidates[i-1])){
continue;
}
cur.push_back(candidates[i]);
if(candidates[i] == target){
res.push_back(cur);
}
else trace(candidates, target-candidates[i], cur, i);
cur.pop_back();
}
return;
}
};