Problem: k Sum (LintCode)
思路
可以把f[i][j][t]
看作表示从前i
个元素中取j
个元素,使其和为t
。
public class Solution {
/**
* @param A: an integer array.
* @param k: a positive integer (k <= length(A))
* @param target: a 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; 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];
}
}
}
return f[n][k][target];
}
}
易错点
初始条件
int[][][] f = new int[n + 1][k + 1][target + 1]; for (int i = 0; i <= n; i++) { f[i][0][0] = 1; }
从
i
个元素里取0个元素,使其和为0的方法就是一种:不取。核心部分
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];
先赋值,再update
f[i][j][t] = 0;
忘掉了先赋值
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