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] ]
Related Topics:
Array, Backtracking
Similar Questions:
// OJ: https://leetcode.com/problems/combination-sum-ii/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
vector<vector<int>> ans;
void dfs(vector<int> &A, int target, int start, vector<int> &tmp) {
if (!target) {
ans.push_back(tmp);
return;
}
for (int i = start; i < A.size() && target >= A[i]; ++i) {
if (i != start && A[i] == A[i - 1]) continue;
tmp.push_back(A[i]);
dfs(A, target - A[i], i + 1, tmp);
tmp.pop_back();
}
}
public:
vector<vector<int>> combinationSum2(vector<int>& A, int target) {
sort(A.begin(), A.end());
vector<int> tmp;
dfs(A, target, 0, tmp);
return ans;
}
};