Lintcode: k Sum

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.

当然这道题可以用Recursion的方法，找出所有满足条件的组合，然后求结果arraylist的size，不过对于只需要求多少个可行解的这道题，找出所有满足条件的组合显得过于奢侈了

如果只需要求一个可行解个数，一个3维DP就好了

没有管优化：要注意base case的选取。

我第一次做的时候只定义了 res[0][0][0] = 1, 其实res[i][0][0] = 1, 漏了这些case，所以当时老是过不了

 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         // write your code here

10         int[][][] res = new int[A.length+1][k+1][target+1];

11         for (int i=0; i<=A.length; i++) {

12             res[i][0][0] = 1;

13         }

14         for (int i=1; i<=A.length; i++) {

15             for (int j=1; j<=k; j++) {

16                 for (int t=1; t<=target; t++) {

17                     if (j > i) res[i][j][t] = 0;

18                     else res[i][j][t] = res[i-1][j][t];

19                     if (t >= A[i-1])

20                         res[i][j][t] += res[i-1][j-1][t-A[i-1]];

21                     

22                 }

23             }

24         }

25         return res[A.length][k][target];

26     }

27 }