[Swift]LeetCode629. K个逆序对数组 | K Inverse Pairs Array

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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 jthelement 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 be 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:

1. The integer n is in the range [1, 1000] and k is in the range [0, 1000].

---

给出两个整数 n 和 k，找出所有包含从 1 到 n 的数字，且恰好拥有 k 个逆序对的不同的数组的个数。

逆序对的定义如下：对于数组的第i个和第 j个元素，如果满i < j且 a[i] > a[j]，则其为一个逆序对；否则不是。

由于答案可能很大，只需要返回 答案 mod 109 + 7 的值。

示例 1:

输入: n = 3, k = 0
输出: 1
解释: 
只有数组 [1,2,3] 包含了从1到3的整数并且正好拥有 0 个逆序对。

示例 2:

输入: n = 3, k = 1
输出: 2
解释: 
数组 [1,3,2] 和 [2,1,3] 都有 1 个逆序对。

说明:

1.  n 的范围是 [1, 1000] 并且 k 的范围是 [0, 1000]。

---

Runtime: 280 ms

Memory Usage: 25.2 MB

class Solution {
    func kInversePairs(_ n: Int, _ k: Int) -> Int {
        var M:Int = 1000000007
        var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:k + 1),count:n + 1)
        dp[0][0] = 1
        if n >= 1
        {
            for i in 1...n
            {
                dp[i][0] = 1
                if k >= 1
                {
                    for j in 1...k
                    {
                        dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % M
                        if j >= i
                        {
                            dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + M) % M
                        }
                    }
                }
            }
        }        
        return dp[n][k]
    }
}