本文主要是介绍[LeetCode]40.Combination Sum II,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
【题目】
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】
基本思路是先排好序,然后每次递归中把剩下的元素一一加到结果集合中,并且把目标减去加入的元素,然后把剩下元素(包括当前加入的元素)放到下一层递归中解决子问题。以start记录我们选到了第几个值,并且一直往后选,这样可以避免选到重复的子集。
这一题和上一题:[LeetCode]39.Combination Sum 差不多一样的思路,只要稍作修改就行。只要在for循环中加一个判断条件即可:
if(i > start && candidates[i] == candidates[i-1]){
continue;
}//if
这个判断的目的是排除同一层次相同元素的出现。例如:下面例题中有两个1,在第一次递归中不能都出现1,可以的是第一次递归出现1,第二次递归也可以出现一个1
【代码】
/*********************************
* 日期:2015-01-27
* 作者:SJF0115
* 题目: 40.Combination Sum II
* 网址:https://oj.leetcode.com/problems/combination-sum-ii/
* 结果:AC
* 来源:LeetCode
* 博客:
**********************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;class Solution {
public:vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {// 中间结果vector<int> path;// 最终结果vector<vector<int> > result;int size = candidates.size();if(size <= 0){return result;}//if// 排序sort(candidates.begin(),candidates.end());// 递归DFS(candidates,target,0,path,result);return result;}
private:void DFS(vector<int> &candidates, int target,int start,vector<int> &path,vector<vector<int> > &result){int len = candidates.size();// 找到一组组合和为targetif(target == 0){result.push_back(path);return;}//iffor(int i = start;i < len;++i){// 同一层次不能出现相同元素if(i > start && candidates[i] == candidates[i-1]){continue;}//if// 剪枝if(target < candidates[i]){return;}//ifpath.push_back(candidates[i]);DFS(candidates,target-candidates[i],i+1,path,result);path.pop_back();}//for}
};int main(){Solution solution;int target = 8;vector<int> vec;vec.push_back(10);vec.push_back(1);vec.push_back(2);vec.push_back(7);vec.push_back(6);vec.push_back(1);vec.push_back(5);vector<vector<int> > result = solution.combinationSum2(vec,target);// 输出for(int i = 0;i < result.size();++i){for(int j = 0;j < result[i].size();++j){cout<<result[i][j]<<" ";}cout<<endl;}return 0;
}
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