Combination Sum II
Total Accepted: 13710 Total
Submissions: 55908My Submissions
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]
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; }
时间: 2024-11-09 00:49:52