LeetCode 493. Reverse Pairs 归并排序坑 mergesort

重要逆序对计数

最新推荐文章于 2025-01-12 16:08:57 发布

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本文介绍了一种算法，用于解决在一个整数数组中找出所有重要逆序对的问题。重要逆序对定义为：如果i<j且nums[i]>2*nums[j]，则(i,j)构成一个重要逆序对。文章通过递归分解和排序策略，实现了一个高效解决方案。

Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].

You need to return the number of important reverse pairs in the given array.

Example1:

Input: [1,3,2,3,1]
Output: 2

Example2:

Input: [2,4,3,5,1]
Output: 3

Note:

The length of the given array will not exceed 50,000.
All the numbers in the input array are in the range of 32-bit integer.

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The key for this problem is to seperate counting and sorting into 2 diff stages:

class Solution:
    def reversePairs(self, nums):
        l = len(nums)

        def dfs(nums, lo, hi):
            if (lo >= hi):
                return 0
            mid = lo + ((hi - lo) >> 1)
            p1 = dfs(nums, lo, mid)
            p2 = dfs(nums, mid+1, hi)
            p3 = 0
            i, j = mid, hi
            while (i >= lo and j >= mid+1):
                if (nums[i] > nums[j]*2):
                    p3 += (j-mid)
                    i -= 1
                else:
                    j -= 1

            nums[lo:hi + 1] = sorted(nums[lo:hi + 1])
            return p1 + p2 + p3

        return dfs(nums, 0, l - 1)