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 }