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设需要排序的序列为:
int arr[]={1, 4, 8, 2, 55, 3, 4, 8, 6, 4, 0, 11, 34, 90, 23, 54, 77, 9, 2, 9, 4, 10};
核心:在快排时候,我踩了一个坑,搞了两个晚上才看出来坑是什么。下面是实际推演:
{1, 4, 8, 2, 55, 3, 4, 8, 6, 4, 0, 11, 34, 90, 23, 54, 77, 9, 2, 9, 4, 10}
1.
{0, 1, | 8, 2, 55, 3, 4, 8, 6, 4, 4, 11, 34, 90, 23, 54, 77, 9, 2, 9, 4, 10}
2.
{0, 1, |4, 2, 2, 3, 4, 8, 6, 4, 4,|| 8,|| 34, 90, 23, 54, 77, 9, 11, 9, 55, 10}
3.
{0, 1, |3, 2, 2,||| 4,|| 4, 8, 6, 4, 4,|| 8,|| 10, 9, 23, 11, 9,||| 34,||| 77, 54, 55, 90}
4.
{0, 1, |2, 2, 3,||| 4,|| 4,|||| 8, 6, 4, 4,|| 8,|| 9, 9, 10, 11, 23,||| 34,||| 55, 54,|||| 77,|||| 90}
5.
{0, 1, |2, 2, 3,||| 4,|| 4,|||| 4, 6, 4,||||| 8,|| 8,|| 9, 9, 10, 11, 23,||| 34,||| 54, 55,|||| 77,|||| 90}
6.
{0, 1, |2, 2, 3,||| 4,|| 4,|||| 4,|||||| 6, 4,||||| 8,|| 8,|| 9, 9, 10, 11, 23,||| 34,||| 54, 55,|||| 77,|||| 90}
7.
{0, 1, |2, 2, 3,||| 4,|| 4,|||| 4,|||||| 4, 6,||||| 8,|| 8,|| 9, 9, 10, 11, 23,||| 34,||| 54, 55,|||| 77,|||| 90}
核心部分如下图所示:
图中,如果的4居于左侧边缘,则做partition(切分,二分)时,需要判断边缘,如果边缘与切割线在一条线上,则对左侧不操作。同样也需要对右侧进行独立判断。代码如下。。
#include<stdio.h>
void quickSort(int left, int right, int*arr);
int main(){
int arr[]={1, 4, 8, 2, 55, 3, 4, 8, 6, 4, 0, 11, 34, 90, 23, 54, 77, 9, 2, 9, 4, 10};
int size = sizeof(arr)/sizeof(arr[0]);
printf("%d\n",size);
for(int i=0;i<size;i++){
printf("%d\t",arr[i]);
}
printf("\n");
quickSort(0, size-1, arr);
for(int i=0;i<size;i++){
printf("%d\t",arr[i]);
}
printf("\n");
return 0;
}
void quickSort(int left, int right, int* arr){
// printf("%d\t%d",left,right);
if(left>=right)
return;
int left_cur=left;
int right_cur=right;
int temp = arr[left];
while(left<right){
while(arr[right]>=temp && left <right) right--;
if(arr[right]<temp && right > left){
arr[left]=arr[right];
left++;
}
while(arr[left]<=temp && left<right) left++;
if(arr[left]>temp && right > left){
arr[right]=arr[left];
right--;
}
}
//for left;
arr[left]=temp;
if(left>left_cur){
quickSort(left_cur,left-1,arr );
}
if(right<right_cur){
quickSort(left+1,right_cur,arr);
}
return;
}