k Sum

Description

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

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：target,Sum,Explanation,Input,integer,Example,dp
来源： https://www.cnblogs.com/FLAGyuri/p/12078401.html