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LintCode-K Sum

时间:2014-12-24 13:22:01      阅读:206      评论:0      收藏:0      [点我收藏+]

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Given n distinct positive integers, integer k (k <= n) and a number target.

Find k numbers where sum is target. Calculate how many solutions there are?

Example

Given [1,2,3,4], k=2, target=5. There are 2 solutions:

[1,4] and [2,3], return 2.

Analysis:

DP. d[i][j][v] means the way of selecting i elements from the first j elements so that their sum equals to k. Then we have:

d[i][j][v] = d[i-1][j-1][v-A[j-1]] + d[i][j-1][v]

It means two operations, select the jth element and not select the jth element.

Solution:

 

 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 an integer
 7      */
 8     public int  kSum(int A[], int k, int target) {
 9         if (A.length<k) return 0;
10         int[][][] d = new int[k+1][A.length+1][target+1];
11         for (int i=1;i<=A.length;i++)
12             if (A[i-1]<=target){
13                 for (int j=i;j<=A.length;j++)
14                     d[1][j][A[i-1]] = 1;
15             }
16 
17         for (int i=2;i<=k;i++)
18             for (int j=i;j<=A.length;j++)
19                 for (int v = 1; v<=target;v++){
20                     d[i][j][v]=0;
21                     if (j>i) d[i][j][v] += d[i][j-1][v];
22                     if (v>=A[j-1]) d[i][j][v] += d[i-1][j-1][v-A[j-1]];
23                 }
24 
25         return d[k][A.length][target];
26                 
27     }
28 }

 

 

 

LintCode-K Sum

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原文地址:http://www.cnblogs.com/lishiblog/p/4182145.html

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