[LeetCode]493. Reverse Pairs 深入浅出算法讲解和代码示例

1、汇总概要

本提解题思路涵盖了插入排序、二分查找、mergeSort思路。

2、题目

Given an array nums, we call (i, j) an important reverse pair if i < j andnums[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.

3、审题

给一个序列，找出它的反转对个数。

判断反转对规则：if i < j and nums[i] > 2*nums[j].

4、解题思路

最基本的思路是双重循环遍历，时间复杂度是O(n*n)，舍弃。

考虑O(nlogn)的复杂度。

1. 将列表数组分为左右两个子数组，list = [[x,x,x][x,x,x]]，满足左边的数组有序，右边的数组无序；

2. 左边数组元素的index都小于右边数组的index，在遍历右边数组元素时，找到左边数组中比它大2倍的数，即是反转对。

讲到这里大家理解了，其实就是插入排序、二分查找反转对。

初始是左边数组为空，右边数组为原list，遍历右边数组中元素，依次往左边数组中插入排序，二分查找插入的位置。

该算法基本没有额外空间消耗，用的mergeSort性质。

5、代码示例 - Java

public class ReversePairs {public int midSearch(int[] list,int low,int high, int value) throws InterruptedException{int mid = (low+high)/2;if(value>list[mid]){mid++;if(mid<high){midSearch(list,mid,high,value);}else{return mid;}}else if(value<list[mid]){mid--;if(mid>low){midSearch(list,low,mid,value);}else{return mid;}}return mid;}public int insertAndPairs(int[] list,int start, int end,int value, int sum){int j=0;for(j=end+1;j>start;j--){list[j]=list[j-1];//insert the least one into list headif(list[j]>2*value){//find pairssum ++;}}list[start] = value;return sum;}public int getPairs(int[] list) throws InterruptedException{int low, mid, high;int i,j;int len = list.length;int sum = 0;for(i=1;i<len;i++){//merge sortint cur = list[i];low = 0;high = i-1;mid = (low+high)/2;if(cur <= list[0]){//[0,i-1][i,len-1],previous list is in ordersum = insertAndPairs(list,0,i-1,cur,sum);list[0] = cur;}else if(cur>=list[i-1]){//no need actioncontinue; }else{ //mid searchint index = midSearch(list,low,high,cur);//compare all values after index, [index,high]sum = insertAndPairs(list,index,i-1,cur,sum);}}return sum;}public static void main(String[] args) throws InterruptedException{int [] list = {2,4,3,5,1};ReversePairs rp = new ReversePairs();int sum = rp.getPairs(list);System.out.println("\n\nRES: "+sum);}}

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