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Lintcode: k Sum II

时间:2015-03-05 08:04:51      阅读:171      评论:0      收藏:0      [点我收藏+]

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Given n unique integers, number k (1<=k<=n)  and target. Find all possible k integers where their sum is target.

Example
Given [1,2,3,4], k=2, target=5, [1,4] and [2,3] are possible solutions.

这道题同Combination Sum II

 1 public class Solution {
 2     /**
 3      * @param A: an integer array.
 4      * @param k: a positive integer (k <= length(A))
 5      * @param target: a integer
 6      * @return a list of lists of integer 
 7      */ 
 8     public ArrayList<ArrayList<Integer>> kSumII(int A[], int k, int target) {
 9         // write your code here
10         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
11         ArrayList<Integer> path = new ArrayList<Integer>();
12         helper(res, path, A, k, target, 0);
13         return res;
14     }
15     
16     public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path, int[] A, int k, int remain, int index) {
17         if (path.size() == k) {
18             if (remain == 0) {
19                 res.add(new ArrayList<Integer>(path));
20             }
21             return;
22         }
23         for (int i=index; i<A.length; i++) {
24             path.add(A[i]);
25             helper(res, path, A, k, remain-A[i], i+1);
26             path.remove(path.size()-1);
27         }
28     }
29 }

 

Lintcode: k Sum II

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原文地址:http://www.cnblogs.com/EdwardLiu/p/4314783.html

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