leetcode 629. K Inverse Pairs Array

629. K Inverse Pairs Array

User Accepted:
81
User Tried:
363
Total Accepted:
83
Total Submissions:
1074
Difficulty:
Medium

Given two integers

n

and

k

, find how many different arrays consist of numbers from

1

to

n

such that there are exactly

k

inverse pairs.

We define an inverse pair as following:For

ith

and

jth

element in the array, if

i

<

j

and

a[i]

>

a[j]

then it's an inverse pair; Otherwise, it's not.

Since the answer may very large, the answer should be modulo 109 + 7.

Example 1:

Input: n = 3, k = 0
Output: 1
Explanation:
Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.

Example 2:

Input: n = 3, k = 1
Output: 2
Explanation:
The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Note:

The integer

n

is in the range [1, 1000] and

k

is in the range [0, 1000].

思路不难想，dp[i][j]表示前i个数字构成j个逆序对的方法数。最朴素的想法是三重循环做，但是可以优化掉最内层对i-1的那层循环。

问题是对边界的处理真的烦，之前的codeM也是，很多题的思路都是对的，但就是边界的问题或者是写法不好导致题目过不去，总是差那么一点。。。真的伤心。。。

public class Solution {
public int kInversePairs(int n, int k) {
long[][] dp = new long[n+1][k+1];

for(int i=0; i<=n; i++)
dp[i][0] = 1;

for(int i=1; i<=n; i++)
for(int j=1; j<=k; j++) {
dp[i][j] = dp[i][j-1]+dp[i-1][j];
if(j-i >= 0) dp[i][j] -= dp[i-1][j-i];
dp[i][j] += 1000000007;
dp[i][j] %= 1000000007;
}

return (int) dp
[k];
}
}