package com.jiucaiyuan.net.question;
/**
* <pre>
* 一个数组arr中,如果左边的数,比右边的数大,则这两个数组成一个逆序对(不一定相邻),找出数组中所有逆序对
* 采用归并排序的思路进行改进实现
* 时间复杂度O(N*logN)
* </pre>
*
* @Author jiucaiyuan 2022/3/30 23:01
* @mail services@jiucaiyuan.net
*/
public class FindReversePair {
public static void reversePair(int[] arr) {
if (arr == null || arr.length < 2) {
return;
}
process(arr, 0, arr.length - 1);
}
/**
* @param arr
* @param l
* @param r
* @return
*/
public static void process(int[] arr, int l, int r) {
if (l == r) {
return;
}
int mid = l + ((r - l) >> 1);
process(arr, l, mid);
process(arr, mid + 1, r);
merge(arr, l, mid, r);
}
/*
* 合并arr[l,mid] 和 arr[mid+1,r] 两个数组
*/
private static void merge(int[] arr, int l, int mid, int r) {
// System.out.println("合并 "+print(arr,l,mid)+" "+print(arr,mid+1,r));
//合并后的数组
int[] temp = new int[r - l + 1];
int i = 0;
int p1 = l;
int p2 = mid + 1;
//两个数组都有数时
while (p1 <= mid && p2 <= r) {
//先拷贝大的数字,再拷贝小的
if (arr[p1] <= arr[p2]) {
temp[i++] = arr[p2++];
} else {
System.out.println(arr[p1]+">"+arr[p2]);
temp[i++] = arr[p1++];
}
}
//前半段数组长
while (p1 <= mid) {
temp[i++] = arr[p1++];
}
//后半段数组长
while (p2 <= r) {
temp[i++] = arr[p2++];
}
//把合并后的temp数组复制到arr中
for (int index = 0; index < temp.length; index++) {
arr[l + index] = temp[index];
}
}
public static void main(String[] args) {
int[] arr = {3, 4, 2, 5, 1};
System.out.println(print(arr,0,arr.length-1));
reversePair(arr);
}
private static String print(int[] arr, int l, int mid) {
StringBuffer sb = new StringBuffer();
for(int i=l;i<=mid;i++){
sb.append(arr[i]).append(",");
}
return sb.substring(0,sb.length()-1);
}
}