2021-03-20 08:44:27 +00:00
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#### [leetcode 493 翻转对](https://leetcode-cn.com/problems/reverse-pairs/)
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2021-03-20 07:58:25 +00:00
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2021-03-20 08:30:29 +00:00
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> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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2021-03-20 07:58:25 +00:00
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**题目描述**
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给定一个数组 nums ，如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。
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你需要返回给定数组中的重要翻转对的数量。
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示例 1:
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> 输入: [1,3,2,3,1]
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> 输出: 2
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示例 2:
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> 输入: [2,4,3,5,1]
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> 输出: 3
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**题目解析**
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我们理解了逆序对的含义之后，题目理解起来完全没有压力的，这个题目第一想法可能就是用暴力法解决，但是会超时，所以我们有没有办法利用归并排序来完成呢？
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我们继续回顾一下归并排序的归并过程，两个小集合是有序的，然后我们需要将小集合归并到大集合中，则我们完全可以在归并之前，先统计一下翻转对的个数，然后再进行归并，则最后排序完成之后自然也就得出了翻转对的个数。具体过程见下图。
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此时我们发现 6 > 2 * 2，所以此时是符合情况的，因为小数组是单调递增的，所以 6 后面的元素都符合条件，所以我们 count += mid - temp1 + 1；则我们需要移动紫色指针，判断后面是否还存在符合条件的情况。
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我们此时发现 6 = 3 * 2，不符合情况，因为小数组都是完全有序的，所以我们可以移动红色指针，看下后面的数有没有符合条件的情况。这样我们就可以得到翻转对的数目啦。下面我们直接看动图加深下印象吧！
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是不是很容易理解啊，那我们直接看代码吧，仅仅是在归并排序的基础上加了几行代码。
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2021-07-05 14:25:41 +00:00
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Java Code:
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2021-03-20 07:58:25 +00:00
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```java
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class Solution {
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private int count;
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public int reversePairs(int[] nums) {
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count = 0;
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merge(nums, 0, nums.length - 1);
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return count;
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}
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public void merge(int[] nums, int left, int right) {
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if (left < right) {
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int mid = left + ((right - left) >> 1);
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merge(nums, left, mid);
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merge(nums, mid + 1, right);
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mergeSort(nums, left, mid, right);
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}
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}
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public void mergeSort(int[] nums, int left, int mid, int right) {
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int[] temparr = new int[right - left + 1];
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int temp1 = left, temp2 = mid + 1, index = 0;
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//计算翻转对
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while (temp1 <= mid && temp2 <= right) {
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//这里需要防止溢出
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if (nums[temp1] > 2 * (long) nums[temp2]) {
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count += mid - temp1 + 1;
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temp2++;
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} else {
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temp1++;
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}
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}
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//记得归位，我们还要继续使用
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temp1 = left;
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temp2 = mid + 1;
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//归并排序
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while (temp1 <= mid && temp2 <= right) {
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if (nums[temp1] <= nums[temp2]) {
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temparr[index++] = nums[temp1++];
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} else {
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temparr[index++] = nums[temp2++];
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}
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}
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//照旧
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if (temp1 <= mid) System.arraycopy(nums, temp1, temparr, index, mid - temp1 + 1);
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if (temp2 <= right) System.arraycopy(nums, temp2, temparr, index, right - temp2 + 1);
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System.arraycopy(temparr, 0, nums, left, right - left + 1);
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}
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}
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```
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2021-07-05 14:25:41 +00:00
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Python Code:
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```python
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from typing import List
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class Solution:
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count = 0
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def reversePairs(self, nums: List[int])->int:
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self.count = 0
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self.merge(nums, 0, len(nums) - 1)
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return self.count
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def merge(self, nums: List[int], left: int, right: int):
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if left < right:
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mid = left + ((right - left) >> 1)
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self.merge(nums, left, mid)
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self.merge(nums, mid + 1, right)
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self.mergeSort(nums, left, mid, right)
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def mergeSort(self, nums: List[int], left: int, mid: int, right: int):
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temparr = [0] * (right - left + 1)
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temp1 = left
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temp2 = mid + 1
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index = 0
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while temp1 <= mid and temp2 <= right:
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# 这里需要防止溢出
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if nums[temp1] > 2 * nums[temp2]:
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self.count += mid - temp1 + 1
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temp2 += 1
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else:
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temp1 += 1
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# 记得归位，我们还要继续使用
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temp1 = left
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temp2 = mid + 1
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# 归并排序
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while temp1 <= mid and temp2 <= right:
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if nums[temp1] <= nums[temp2]:
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temparr[index] = nums[temp1]
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index += 1
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temp1 += 1
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else:
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temparr[index] = nums[temp2]
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index += 1
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temp2 += 1
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# 照旧
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if temp1 <= mid:
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temparr[index: index + mid - temp1 + 1] = nums[temp1: temp1 + mid - temp1 + 1]
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if temp2 <= right:
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temparr[index: index + right - temp2 + 1] = nums[temp2: temp2 + right - temp2 + 1]
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nums[left: left + right- left + 1] = temparr[0: right - left + 1]
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```
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