首页 > 代码库 > 【LeetCode】Combination Sum
【LeetCode】Combination Sum
题目
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
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 2,3,6,7
and target 7
,
A solution set is: [7]
[2, 2, 3]
题目给出一个候选数列和一个目标数值,找出所有相加等于目标数值的数列,原候选数列的元素可以重复。
NP问题,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。代码如下:
public class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> res=new ArrayList<List<Integer>>(); //注意多维List的初始化 if(candidates==null||candidates.length==0){ return res; } Arrays.sort(candidates); helper(candidates,0,target,new ArrayList<Integer>(),res); return res; } private void helper(int[] candidates,int start,int target,ArrayList<Integer> item,List<List<Integer>> res){ //注意和前面类型对应 if(target<0){ return; } if(target==0){ res.add(new ArrayList<Integer>(item)); return; } for(int i=start;i<candidates.length;i++){ if(i>0&&candidates[i]==candidates[i-1]){ continue; } item.add(candidates[i]); helper(candidates,i,target-candidates[i],item,res); item.remove(item.size()-1); } } }
参考:http://blog.csdn.net/linhuanmars/article/details/20828631
---EOF---
【LeetCode】Combination Sum
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。