leetcode 040 Combination Sum II

最新推荐文章于 2019-02-21 08:43:58 发布

原创最新推荐文章于 2019-02-21 08:43:58 发布 · 310 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

127 篇文章

订阅专栏

本文探讨了给定一组候选数和目标数时如何找出所有不重复的组合，使得组合内的数相加等于目标数。文章提供了详细的算法实现，并强调了元素必须按非递减顺序排列且每个数只能使用一次。

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]

Subscribe to see which companies asked this question

class Solution {
public:
    void findNum(vector<int> &candinates, vector<int> container, vector<vector<int> > &ans, int target, int index, int sum, int n, int flag) {
		if(sum == target) {
			ans.push_back(container);
			return;
		}

		if(index >= n || sum > target || sum+candinates[index] > target) return;
        
        if(index-1 >= 0 && candinates[index]==candinates[index-1] && flag==true || index==0 || candinates[index]!=candinates[index-1]) {
		    container.push_back(candinates[index]);
	        findNum(candinates, container, ans, target, index+1, sum+candinates[index], n, true);
	        container.erase(container.end()-1);
		    findNum(candinates, container, ans, target, index+1, sum, n, false);
        } else {
            findNum(candinates, container, ans, target, index+1, sum, n, false);
        }
        
		
	}
	vector<vector<int> > combinationSum2(vector<int> &candinates, int target) {
		sort(candinates.begin(), candinates.end());
		vector<vector<int> > ans;
		vector<int> container;

		findNum(candinates, container, ans, target, 0, 0, candinates.size(), false);

		return ans;
	}
};