K个逆序对数组

public class Solution {
public int kInversePairs(int n, int k) {
int[][] dp = new int[n + 1][k + 1];
int M = 1000000007;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= k && j <= i * (i - 1) / 2; j++) {
if (i == 1 && j == 0) {
dp[i][j] = 1;
break;
} else if (j == 0)
dp[i][j] = 1;
else {
int val = (dp[i - 1][j] + M - ((j - i) >= 0 ? dp[i - 1][j - i] : 0)) % M;
dp[i][j] = (dp[i][j - 1] + val) % M;
}
}
}
return dp[n][k];
}
}

作者：LeetCode
链接：https://leetcode-cn.com/problems/k-inverse-pairs-array/solution/kge-ni-xu-dui-shu-zu-by-leetcode/
来源：力扣（LeetCode）
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