标签:pre 快速排序 rgs count [] string 数组 merge 好的
tag: 算法基本功 -> 排序 ->快速排序
归并排序思路:
(1) 通过递归的方式把数据拆分成两两数据对(可能存在落单的数)
(2) 将两两数据对进行排序同时将排序结果存在临时数组中
(3) 把临时数组中已经排好的数按照对应的序号更新到原数组中
package com.zhaochao.sort;
/**
* Created by zhaochao on 17/1/16.
*/
public class MergeSort {
public void mergeSort(int[] arrs, int left, int right) {
if(left >= right || arrs == null || arrs.length <= 1) {
return;
}
int mid = (left + right) / 2;
mergeSort(arrs, left, mid);
mergeSort(arrs,mid + 1, right);
sort(arrs,left,right);
}
public void sort(int[] arrs, int left, int right) {
if(left >= right) {
return;
}
int mid = (left + right) / 2;
int[] tmp = new int[right - left + 1];
int i = left;
int j = mid + 1;
int count = 0;
while((i <= mid) && (j <= right)) {
if ( arrs[i] < arrs[j]) {
tmp[count++] = arrs[i++];
}
else {
tmp[count++] = arrs[j++];
}
}
while(i <= mid){
tmp[count++] = arrs[i++];
}
while(j <= right) {
tmp[count++] = arrs[j++];
}
for(int p = 0; p < right - left + 1; p++) {
arrs[p + left] = tmp[p];
}
}
public static void main(String[] args) {
MergeSort test = new MergeSort();
//int[] arrs = {9,8,7,6,1,2,3};
int[] arrs = {9,8,7,6,1,2};
test.mergeSort(arrs,0,5);
for(int i = 0; i < 6; i++) {
System.out.println(arrs[i]);
}
}
}
标签:pre 快速排序 rgs count [] string 数组 merge 好的
原文地址:http://www.cnblogs.com/superzhaochao/p/6312583.html