1 //将一个字符串数组的元素的顺序进行反转。{"3","a","8","haha"} {"haha","8","a","3"}。第i个和第length-i-1个进行交换。
2 string[] strs = { "3", "a", "8", "haha" };
3 for (int i = 0; i < strs.Length / 2; i++)
4 {
5 //i,len-1-i
6 //交换两个元素,想像一下交换i,j,交换两个杯子中的水
7 string temp = strs[i];
8 strs[i] = strs[strs.Length - 1 - i];
9 strs[strs.Length - 1 - i] = temp;
10 }
11
12 for (int i = 0; i < strs.Length; i++)
13 {
14 Console.Write(strs[i]+",");
15 }
2014年6月7日03:49:28,布布扣,bubuko.com
时间: 2024-10-12 08:11:21