Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, � , ak) must be in non-descending order. (ie, a1 ? a2 ? � ? ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5
and target 8
,
A solution set is: [1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
思路:
同样DFS,跟上一题是一样的,只不过为了不重复,每次调用dfs的时候pos = i + 1;
这道题备选数组是有重复的,得去重
代码:
class Solution { public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<int> temp; vector<vector<int> > rst; sort(num.begin(), num.end()); dfs(num, target, temp, rst, 0); return rst; } void dfs(vector<int> &num, int target, vector<int> &temp, vector<vector<int> > &rst, int pos) { if (target < 0) return; if (target == 0) { rst.push_back(temp); return; } for (int i = pos; i < num.size(); i++) { temp.push_back(num[i]); dfs(num, target - num[i], temp, rst, i + 1); temp.pop_back(); while (i < num.size() - 1 && num[i] == num[i+1]) i++; } } };