Leetcode 40 - 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]

思路：与Leetcode 39思路大致相同，依然使用深度优先搜索。本题还应考虑如何防止产生重复解。

class Solution {
public:
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
sort(num.begin(),num.end());
vector<int> temp;
vector<vector<int>> result;
dfs(num,target,0,temp,result);
return result;
}

private:
void dfs(vector<int> &num,int gap,int start,vector<int> &temp,vector<vector<int>> &result){
if(gap==0)
{
result.push_back(temp);
return;
}

int previous = -1; //记录本次递归选择的数，防止重复
for(int i=start;i<num.size();i++)
{
//防止重复
if(num[i]==previous) continue;
if(gap<num[i]) return;

previous = num[i];

temp.push_back(num[i]);
dfs(num,gap-num[i],i+1,temp,result);
temp.pop_back();
}
}
};