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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] ]
给定一个都是正数的数组，元素可能有重复，让你找出所有和等于给定数的数组。
思路：递归、深度优先搜索、回溯
重复的数可以使用但不能给出重复的组合，需要加一个判断条件
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int> > ans;
vector<int> cur;
sort(candidates.begin(), candidates.end());
dfs(candidates, target, 0, ans, cur);
return ans;
}
void dfs(vector<int>& candidates, int target, int start, vector<vector<int> >& ans, vector<int>& cur) {
if (target < 0) return;//加速计算
if (target == 0) {
ans.push_back(cur);
return;//加速计算，找到一种情况，后面都不可能满足了
}
for (int i = start; i < candidates.size(); i++) {
if (i > start && candidates[i] == candidates[i - 1]) continue;//在一次查找下，重复数字不再使用，以便去掉重复组合
cur.push_back(candidates[i]);
dfs(candidates, target - candidates[i], i + 1, ans, cur);//这里从i+1 开始，而不是start+1,会有重复
cur.pop_back();
}
}
};