Leetcode:Combination Sum II
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]
分析:解题思路跟Combination Sum类似,但在同一层递归时要跳过相同的元素。代码如下:
class Solution { public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int> > result; vector<int> path; sort(num.begin(), num.end()); dfs(result, path, num, 0, target); return result; } void dfs(vector<vector<int> > &result, vector<int> &path, vector<int> &num, int start, int gap){ if(gap == 0){ result.push_back(path); return; } for(int i = start; i < num.size(); i++){ if(gap < num[i]) return; // prune if(i == start || num[i-1] != num[i]){ path.push_back(num[i]); dfs(result, path, num, i+1, gap-num[i]); path.pop_back(); } } } };