【字符串问题】求一个字符串中连续出现次数最多的子串

2013-09-14 10:47:39

在面试宝典上看到的题目，自己做了一下，用了C++中的string类，比较方便。

思路：

1. 遍历源字符串的每一个字符，以该字符为首的重复子串的长度为1到以该字符为首的后缀字符串（即以该字符串为首，到字符结尾的子串，比如abdcef的第三个后缀字符串即为dcef）的长度的一半；
2. 对所有的子串，判断连续重复出现的次数，如果子串连续重复出现的次数大于当前最大连续重复出现的次数，则更新最大连续重复出现的次数。

注意：

VC6.0 对C++的STL支持不是很好，有的方法不支持，如下面代码中的 srcStr.clear();在VS2008中可以运行无措，但在VC6.0中就会报错如下：

 error C2039: 'clear' : is not a member of 'basic_string<char,struct std::char_traits<char>,class std::allocator<char> >'

解决方法如下：

用可以完成相同功能的subStr.erase(0,subStr.length());替代，不会报错。

目前还没发现其他更好的办法，如有发现，还请分享一下哦！

代码如下（测试暂未发现错误，欢迎交流指正）：

#include <iostream>
#include <cassert>
#include <string>

using namespace std;

//对string类型，如何测试输入的有效性
size_t FindSubstrAppearMaxtimes(const string srcStr,string &subStr,size_t &count)
{
    count = 0;
    subStr.clear();  //或subStr.erase(0,subStr.length());
    
    string tmpStr;
    size_t tmpCount = 0;
    string tmpStrToComp;
    size_t posOfSubStr = 0;

    size_t srcEnd = srcStr.length();
    size_t srcCurIndex = 0;
    size_t subLen = 0;
    size_t subCurIndex = 0;

    while (srcCurIndex < srcEnd)
    {
        for (subLen = 1;2 * subLen <= srcEnd - srcCurIndex;++subLen)
        {
            tmpStr = srcStr.substr(srcCurIndex,subLen);
            tmpCount = 0;
            tmpStrToComp.erase(0,tmpStrToComp.length());

            for (subCurIndex = srcCurIndex;subCurIndex < srcEnd;subCurIndex += subLen)
            {
                tmpStrToComp = srcStr.substr(subCurIndex,subLen);
                if (tmpStr == tmpStrToComp)   //要加string头文件，才能用==
                {
                    ++tmpCount;
                }
                else
                {
                    if (tmpCount > count)
                    {
                        posOfSubStr = srcCurIndex;
                        count = tmpCount;
                        subStr = tmpStr;
                    }
                    break;
                }
            }
        }

        ++srcCurIndex;
    }

    return posOfSubStr;
}

void TestDriver()
{
    string strArray[] = {"0123456","abcdef","abcbcbcedhellobc","abcbcbcabc","abcccabc"};
    size_t arrayLength = 5;

    string srcStr;
    string subStr;
    size_t count = 0;
    size_t posOfSubStr = 0;

    for (size_t index = 0;index < arrayLength;++index)
    {
        srcStr.clear();
        srcStr = strArray[index];
        posOfSubStr = FindSubstrAppearMaxtimes(srcStr,subStr,count);

        cout<<"the source string is : "<<srcStr<<endl;
        cout<<"the sub string appear most frequently is : "<<subStr<<endl;
        cout<<"the appearance times is : "<<count<<endl;
        cout<<endl;
    }
}

int main()
{
    TestDriver();
    return 0;
}

测试结果：

the source string is : 0123456
the sub string appear most frequently is : 0
the appearance times is : 1

the source string is : abcdef
the sub string appear most frequently is : a
the appearance times is : 1

the source string is : abcbcbcedhellobc
the sub string appear most frequently is : bc
the appearance times is : 3

the source string is : abcbcbcabc
the sub string appear most frequently is : bc
the appearance times is : 3

the source string is : abcccabc
the sub string appear most frequently is : c
the appearance times is : 3

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