基于visual Studio2013解决面试题之1102合并字符串

来源：互联网发布：array chunk php 编辑：程序博客网时间：2024/05/07 09:32

题目

解决代码及点评

/*    给出一个函数来合并两个字符串 A 和 B 。字符串 A 的后几个字节和字符串 B 的前几个字节重叠重叠部分只要一份*/#include <iostream>using namespace std;void CopyStr(char *pszBuf1, char *pszBuf2, char *pszBuf3){    int nLen1 =strlen(pszBuf1);    int nLen2 =strlen(pszBuf2);    int nLen3 =nLen1 + nLen2 + 1;    int nPos1 = 0;    int nPos2 = 0;    int nPos3 = 0;    while (nPos1 < nLen1 - 1 && nPos2 < nLen2 - 1 && nPos3 < nLen3 - 1)    {        while (pszBuf1[nPos1] != pszBuf2[0] && nPos1 < nLen1 - 1)        {            pszBuf3[nPos3++] = pszBuf1[nPos1++];        }        if (nPos1 == nLen1 - 1)        {//pszBuf1完全结束，不存在重复            pszBuf3[nPos3++] = pszBuf2[nPos2++];        }        else        {//            int nTmpPos = nPos1;            while (pszBuf1[nPos1] == pszBuf2[nPos2] && (nPos1 < nLen1) && (nPos2 < nLen2))            {                nPos2++;                nPos1++;            }            if (nPos1 == nLen1)            {//完全重叠                nPos2 = 0;                while (nPos2 < nLen2)                {                    pszBuf3[nPos3++] = pszBuf2[nPos2++];                }            }            else            {//不完全重叠，回复nPos1                nPos1 = nTmpPos;                nPos2 = 0;                pszBuf3[nPos3++] = pszBuf1[nPos1++];            }        }    }        }int main(){    char szBuf1[] = "aghgh";    char szBuf2[] = "ghjklqwe";    int nLen3 = strlen(szBuf1) + strlen(szBuf2) + 1;    char *pszResult = new char[nLen3];int i;    for (i = 0; i < nLen3; i++)    {        pszResult[i] = 0;    }    pszResult[i-1] = '\0';    CopyStr(szBuf1, szBuf2, pszResult);    cout<<pszResult<<endl;    system("pause");    return 0;}

代码下载及其运行

代码下载地址：http://download.csdn.net/detail/yincheng01/6704519

解压密码：c.itcast.cn

下载代码并解压后，用VC2013打开interview.sln，并设置对应的启动项目后，点击运行即可，具体步骤如下：

1）设置启动项目：右键点击解决方案，在弹出菜单中选择“设置启动项目”

2）在下拉框中选择相应项目，项目名和博客编号一致

3）点击“本地Windows调试器”运行

程序运行结果

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