回顾冒泡排序(新增优化代码)

//
//  main.c
//  数组-冒泡排序
//
//  Created by LongMa on 2019/6/26.
//  Copyright © 2019. All rights reserved.
//
 
#include <stdio.h>

int main(int argc, const char * argv[]) {
    int a[10];
    for (int i = 0; i < 10; i++) {
        printf("请输入第%d个整数,共10个:",i + 1);
        scanf("%d", &a[i]);
    }
     
    int temp = 0;
     
    //冒泡：第一次，最大的排到了最后一位；第二次（不用对比最后一个），次大的排到倒数第二位...
    for (int i = 0; i <= 10 - 1 - 1 ; i++) {//共3个数，要循环2次 => n个数，循环n-1次
        int endLoop = 0;
        for (int j = 0; j <= 10 - 1 - 1 - i ; j++) {//i为0时，j + 1最大 == 9,j最大 == 8
            if (a[j] > a[j + 1]) {
                temp = a[j];
                a[j] = a[j + 1];
                a[j + 1] = temp;
                endLoop = 1;
                printf("exchanged
");
            }
        }
        
        printf("第%d次循环
", i);
        //未发生过交换，说明已全部有序。结束循环
        if (0 == endLoop) {
            break;
        }
    }
     
    for (int k = 0; k < 10; k++) {
        printf("%d	",a[k]);
    }
    return 0;
}

 log：

请输入第1个整数,共10个:1
请输入第2个整数,共10个:2
请输入第3个整数,共10个:3
请输入第4个整数,共10个:4
请输入第5个整数,共10个:5
请输入第6个整数,共10个:6
请输入第7个整数,共10个:8
请输入第8个整数,共10个:7
请输入第9个整数,共10个:10
请输入第10个整数,共10个:9
exchanged
exchanged
第0次循环
第1次循环
1    2    3    4    5    6    7    8    9    10    Program ended with exit code: 0

swift实现版

 import Foundation
 
 func mp(arr: Array<Int>) ->Array<Int> {
    var a = arr
    let len = a.count
    for i in 0...(len - 2){
        var endLoop = 0
        for j in 0...(len - 2 - i){
            if a[j] > a[j + 1] {
                
                //exchange
                let temp = a[j + 1]
                a[j + 1] = a[j]
                a[j] = temp
                endLoop = 1
                print("exchanged")
            }
        }
        
        print("第(i)次循环")
        if 0 == endLoop{
            break;
        }
        
    }
    return a
 }
 
 let gArr : [Int] = [6, 3, 5, 4, 88, 9, 2, 1, 0]
 let lRslt = mp(arr: gArr)
 print(lRslt)
 
 print("---------")
 let gArr1 : [Int] = [1, 2, 3, 4, 5, 6, 8, 7, 9]
 let lRslt1 = mp(arr: gArr1)
 print(lRslt1)

log:

exchanged
exchanged
exchanged
exchanged
exchanged
exchanged
exchanged
第0次循环
exchanged
exchanged
exchanged
exchanged
第1次循环
exchanged
exchanged
exchanged
第2次循环
exchanged
exchanged
exchanged
第3次循环
exchanged
exchanged
exchanged
第4次循环
exchanged
exchanged
exchanged
第5次循环
exchanged
exchanged
第6次循环
exchanged
第7次循环
[0, 1, 2, 3, 4, 5, 6, 9, 88]
---------
exchanged
第0次循环
第1次循环
[1, 2, 3, 4, 5, 6, 7, 8, 9]