2.2-2 选择排序

#include <stdio.h>

void main(){
	int i, j, m;
	int min, index;
	int a[6] = {5,2,4,6,1,3};

	for (i = 0; i < 6; i++){
		min = a[i];
		for (j = i+1; j < 6; j++){
			if (a[j] < min) {
				min = a[j];
				index = a[i];
				a[i] = min;
				a[j] = index;
			}
		}
	}
	for (m=0; m<6; m++) {  
        printf("%d ", a[m]);  
    }  
}

以上程序正确输出。

错误程序：

#include <stdio.h>

void main(){
	int i, j, m;
	int min, key, index;
	int a[6] = {5,2,4,6,1,3};

	for (i = 0; i < 6; i++){
        min = a[i];
		for (j = i+1; j < 6; j++){
			if (a[j] < min) {
				min = a[j];
				key = j;
			}
		}
		index = a[i];
		a[i] = min;
		a[key] = index;
	}
	for (m=0; m<6; m++) {  
        printf("%d ", a[m]);  
    }  
}

原因分析：

在这三步中

	index = a[i];
	a[i] = min;
	a[key] = index;

如果此次j循环没有做任何动作，那么此时的key值仍然为上一个i循环中的值，这时的交换出现错误

例如 5 2 4 6 1 3 第一个i循环后，key=4，序列变成 1 2 4 6 5 3 

index = a[1] = 2;

a[1] = 2;

a[4] = 2;

此时错误出现，将 a[4] 原本为5的值擅自改变为2 

但是程序一，交换次数太多。引入key的改进

#include <stdio.h>

void main(){  
    int i, j, m;  
    int min, key, index;  
    int a[6] = {5,2,4,6,1,3};  
  
    for (i = 0; i < 6; i++){  
        min = a[i]; 
	key = i; 
        for (j = i+1; j < 6; j++){  
            if (a[j] < min) {  
                min = a[j];  
                key = j;  
            }  
        }
		if (key > i) { //保证只有需要交换时才交换
			index = a[i];  
			a[i] = min;  
			a[key] = index;  
		}
    }  
    for (m=0; m<6; m++) {    
        printf("%d ", a[m]);    
    }    
}  

第一个关于 i 的 for 循环，i < 5 就行。

应用loop invariant思想，sequence最后一个数肯定是最大的。

As a loop invariant, we choose that A[1...i-1] are sorted and all other elements are greater then these. We only need to iterate to n-1 since according to the invariant the n-th element will be the largest.