标签：des style http color os io ar for art
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:
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]
题意：给定一组数C和一个数值T，在C中找到所有总和等于T的组合。C中的同一数字最多只能拿一次，找到的组合不能重复。
思路：dfs
第i层的第j个节点有 n - i - j 个选择分支
递归深度：递归到总和大于等于T就可以返回了
复杂度：时间O(n!)，空间O(n)
vector<vector<int> > res;
vector<int> _num;
void dfs(int start, int target, vector<int> &path){
if(target == 0) {res.push_back(path); return;}
int previous = -1; //这里要加上这个来记录同一层分枝的前一个值，如果当前值跟前一个值一样，就跳过，避免重复
for(int i = start; i < _num.size(); ++i){
if(previous == _num[i]) continue;
if(target < _num[i]) return; //剪枝
previous = _num[i];
path.push_back(_num[i]);
dfs(i + 1, target - _num[i], path);
path.pop_back();
}
}
vector<vector<int> > combinationSum2(vector<int> &num, int target){
_num = num;
sort(_num.begin(), _num.end());
vector<int> path;
dfs(0, target, path);
return res;
}Leetcode dfs Combination SumII
标签：des style http color os io ar for art
原文地址：http://blog.csdn.net/zhengsenlie/article/details/38979127
