排序

1.冒泡排序

方法1

#include <iostream>
using namespace std;
int main()
{
	int a[10] = { 9,7,6,4,10,8,5,3,1,2 };
	int i, j;
	int t;
	int length = sizeof(a)/sizeof(a[0]); //获取数组长度
	for (i = 0; i <= length - 1; i++)
	{
		for (j = 0; j <= length - 2; j++) //若为length-1数组越界
		{
			if (a[j] > a[j + 1])
			{
				t = a[j + 1];
				a[j + 1] = a[j];
				a[j] = t;
			}
		}
	}
	for (i = 0; i <= length - 1; i++)
	{
		cout << a[i] << endl;
	}
	return 0;
}

方法2(每遍历一遍判断是否排序完成，能够提前结束）

#include <iostream>
using namespace std;
int main()
{
	int a[10] = { 9,7,6,4,10,8,5,3,1,2 };
	int i, j;
	int t;
	int length = sizeof(a)/sizeof(a[0]); //获取数组长度
	bool flag = false;
	for (i = 0; i <= length - 1 && !flag; i++)
	{
		flag = true;
		for (j = 0; j <= length - 2; j++) //若为length-1数组越界
		{
			if (a[j] > a[j + 1])
			{
				t = a[j + 1];
				a[j + 1] = a[j];
				a[j] = t;
				flag = false;
			}
		}
	}
	for (i = 0; i <= length - 1; i++)
	{
		cout << a[i] << endl;
	}
	return 0;
}

方法3(在每次排序后会有最大值出现在该次排序的末尾，此最大值不参与下一次的排序；可能一次会排好最大值和第二大值等，故设置upto=j而不是n--)

#include <iostream>
using namespace std;
int main()
{
	int a[10] = { 9,7,6,4,10,8,5,3,1,2 };
	int i, j;
	int t;
	int length = sizeof(a)/sizeof(a[0]); //获取数组长度
	int n = length - 1;
	int upto = 0;
	for (i = 0; i <= length - 1; i++)
	{
		for (j = 0; j <= n - 1; j++) //若为length数组越界
		{
			if (a[j] > a[j + 1])
			{
				t = a[j + 1];
				a[j + 1] = a[j];
				a[j] = t;
				upto = j;
			}
		}
		n = upto;
	}
	for (i = 0; i <= length - 1; i++)
	{
		cout << a[i] << endl;
	}
	return 0;
}

方法4(每次循环排序好最大值最小值，并判断没有排序号的范围以进行下次循环排序的优化)

#include <iostream>
using namespace std;
int main()
{
	int a[10] = { 9,7,6,4,10,8,5,3,1,2 };
	int i;
	int length = sizeof(a)/sizeof(a[0]); //获取数组长度
	int first = 0;
	int last = length - 1;
	int t;
	while (first < last)
	{
		int nfirst = first, nlast = last;//用于判断循环中有无变换以及未排序好的范围
		for (i = first; i <= last - 1; i++) //正向循环
		{
			if (a[i] > a[i + 1])
			{
				nlast = i;
				t = a[i];
				a[i] = a[i + 1];
				a[i + 1] = t;
			}
		}//将最大值排到最后
		if (nlast == last) //若此次循环没发生变换即排序完成
		{
			break;
		}
		last = nlast;
		for (i = last; i >= first + 1; i--) //逆向循环
		{
			if (a[i] < a[i - 1])
			{
				nfirst = i;
				t = a[i];
				a[i] = a[i - 1];
				a[i - 1] = t;
			}
		}
		if (nfirst == first) //若此次循环没发生变换即排序完成
		{
			break;
		}
		first = nfirst;
	}
	for (i = 0; i <= length - 1; i++)
	{
		cout << a[i] << endl;
	}
	return 0;
}

2.选择排序