★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）➤博客园地址：山青咏芝（）➤GitHub地址：➤原文地址： ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★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 = 0Output: 1Explanation: 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 = 1Output: 2Explanation: The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.
Note:
- The integer
nis in the range [1, 1000] andkis 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:
-
n的范围是 [1, 1000] 并且k的范围是 [0, 1000]。
Runtime: 280 ms
Memory Usage: 25.2 MB
1 class Solution { 2 func kInversePairs(_ n: Int, _ k: Int) -> Int { 3 var M:Int = 1000000007 4 var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:k + 1),count:n + 1) 5 dp[0][0] = 1 6 if n >= 1 7 { 8 for i in 1...n 9 {10 dp[i][0] = 111 if k >= 112 {13 for j in 1...k14 {15 dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % M16 if j >= i17 {18 dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + M) % M19 }20 }21 }22 }23 } 24 return dp[n][k]25 }26 }