标签：wrap lan 动态 tput solution alc The open script
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 1
Input:
List = [1,2,3,4]
k = 2
target = 5
Output: 2
Explanation: 1 + 4 = 2 + 3 = 5
Example 2
Input: List = [1,2,3,4,5] k = 3 target = 6 Output: 1 Explanation: There is only one method. 1 + 2 + 3 = 6思路：采用动态规划（dp）的思想，进行状态转移，记录数字和和出现次数之间的关系。
用dp[i][j][k]dp[i][j][k]表示前ii个数里选jj个和为k的方案数。
public class Solution {
/**
* @param A: An integer array
* @param k: A positive integer (k <= length(A))
* @param target: An integer
* @return: An integer
*/
public int kSum(int[] A, int k, int target) {
int n = A.length;
int[][][] f = new int[n + 1][k + 1][target + 1];
for (int i = 0; i < n + 1; i++) {
f[i][0][0] = 1;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= k && j <= i; j++) {
for (int t = 1; t <= target; t++) {
f[i][j][t] = 0;
if (t >= A[i - 1]) {
f[i][j][t] = f[i - 1][j - 1][t - A[i - 1]];
}
f[i][j][t] += f[i - 1][j][t];
} // for t
} // for j
} // for i
return f[n][k][target];
}
}
标签：wrap lan 动态 tput solution alc The open script
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078401.html
