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>
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#include<malloc.h>
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#include<iostream>
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using namespace std;
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//#define swap(x,y) (x ^= y,y ^= x,x ^= y)
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int minTime(int *a,int n){
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if (a ==NULL || n == 0)
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return 0;
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int i = 0,j = 0;
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int count = 0;
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for (i = 0;i < n-1;i++)
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{
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for (j = 0;j < n-i-1;j++)
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{
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if (a[j] > a[j+1])
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{
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swap(a[j],a[j+1]);
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count++;
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}
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}
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}
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return count;
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}
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int main()
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{
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int len = 0;
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while (cin >> len)
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{
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int *a = (int *)malloc(sizeof(int)*len);
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int i = 0;
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for (i = 0;i < len;i++)
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{
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cin >> a[i];
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}
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int result = minTime(a,len);
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printf("%d\n",result);
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for (i = 0;i < len;i++)
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{
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printf("%d\n",a[i]);
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}
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free(a);
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}
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return 0;
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}
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>
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#include<vector>
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using namespace std;
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class Coder {
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public:
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vector<string> findCoder(vector<string> A, int n) {
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vector<string> result;
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vector<int> resultInt;
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//int count = 0;
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vector<string>::iterator iter = A.begin();
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for (;iter != A.end();iter++)
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{
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int j =0;
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string s = *iter;
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int count = 0;
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for (int i = 0;i < (*iter).length() && i+4 < (*iter).length();)
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{
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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')
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{
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i = i+5;
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count++;
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}
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else
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i = i+1;
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}
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//cout << count << endl;
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result.push_back(*iter);
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resultInt.push_back(count);
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}
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for (int i = 0;i < n-1;i++)
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{
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for (int j = 0;j < n-i-1;j++)
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{
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if (resultInt[j] < resultInt[j+1])
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{
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swap(resultInt[j],resultInt[j+1]);
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swap(result[j],result[j+1]);
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}
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}
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}
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for (int i = 0;i < n;i++)
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{
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cout << result[i]<< endl;
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cout << resultInt[i]<< endl;
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}
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return result;
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}
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};
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int main()
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{
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Coder coder;
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string str[3] = {"i am a coder","Coder,Coder","Coder"};
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vector<string> A(&str[0],&str[3]);
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coder.findCoder(A,3);
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return 0;
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}
运行结果:
[root@bogon ~]# ./a.out
Coder,Coder
2
i am a coder
1
Coder
1
[root@bogon ~]#
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