经典排序算法之冒泡排序

基本思路：比较表中的相邻元素，如果他们是逆序的话就交换他们的位置。重复多次以后，最终最大 的元素就沉到列表的最后一个位置，第二遍操作将第二大的元素沉下去，这样一直做，知道n-1遍以后，该列表就排好序了。 代码如下：

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#include<stdio.h> #include<malloc.h> #include<iostream> using namespace std; //#define swap(x,y) (x ^= y,y ^= x,x ^= y) int minTime(int *a,int n){     if (a ==NULL || n == 0)         return 0;     int i = 0,j = 0;     int count = 0;     for (i = 0;i < n-1;i++)     {         for (j = 0;j < n-i-1;j++)         {             if (a[j] > a[j+1])             {                 swap(a[j],a[j+1]);                 count++;             }                          }     }     return count; } int main() {         int len = 0;    while (cin >> len)    {        int *a = (int *)malloc(sizeof(int)*len);        int i = 0;        for (i = 0;i < len;i++)        {             cin >> a[i];        }        int result = minTime(a,len);        printf("%d\n",result);        for (i = 0;i < len;i++)        {         printf("%d\n",a[i]);     }        free(a);    }     return 0; } 包括比较次数 运行结果如下： [root@localhost ~]# ./a.out 4 4 3 2 1 6 1 2 3 4 去哪笔试题(2016.09.08)

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#include<iostream> #include<vector> using namespace std; class Coder { public:     vector<string> findCoder(vector<string> A, int n) {           vector<string> result;     vector<int> resultInt;     //int count = 0;     vector<string>::iterator iter = A.begin();     for (;iter != A.end();iter++)     {         int j =0;         string s = *iter;         int count = 0;         for (int i = 0;i < (*iter).length() && i+4 < (*iter).length();)         {             if (toupper(s[i]) == 'C' && toupper(s[i+1]) == 'O' && toupper(s[i+2]) == 'D' && toupper(s[i+3]) == 'E' && toupper(s[i+4]) == 'R')             {                 i = i+5;                 count++;             }             else                 i = i+1;         }         //cout << count << endl;         result.push_back(*iter);         resultInt.push_back(count);     }     for (int i = 0;i < n-1;i++)     {         for (int j = 0;j < n-i-1;j++)         {             if (resultInt[j] < resultInt[j+1])             {                 swap(resultInt[j],resultInt[j+1]);                 swap(result[j],result[j+1]);             }         }     }     for (int i = 0;i < n;i++)     {         cout << result[i]<< endl;         cout << resultInt[i]<< endl;     }     return result;     } }; int main() {     Coder coder;     string str[3] = {"i am a coder","Coder,Coder","Coder"};     vector<string> A(&str[0],&str[3]);     coder.findCoder(A,3);     return 0; } 运行结果： [root@bogon ~]# ./a.out Coder,Coder 2 i am a coder 1 Coder 1 [root@bogon ~]#   <script>window._bd_share_config={"common":{"bdSnsKey":{},"bdText":"","bdMini":"2","bdMiniList":false,"bdPic":"","bdStyle":"0","bdSize":"16"},"share":{}};with(document)0[(getElementsByTagName('head')[0]||body).appendChild(createElement('script')).src='http://bdimg.share.baidu.com/static/api/js/share.js?v=89860593.js?cdnversion='+~(-new Date()/36e5)];</script> 阅读(161) | 评论(0) | 转发(0) | 0

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