利用后缀数组求字符串的最长重复子串的算法 利用二维数组求两个字符串的最长公共字串(动态规划)

[问题描述]

 

1-  求一个字符串中存在的最长的重复子串[后缀数组]

2-  求两个字符串的最长公共字串            [动态规划]

 

[问题解决]

 

1-利用  “后缀数组”(需排序)  求字符串中存在的最长的重复子串，要建立指针数组并通过qsort排序

   时间复杂度是O(nlogn)主要开销在于排序的过程。

 

2-利用 “二维bool数组”来解决两个字符串的最长公共子串，找出正对角线中连续true最长的对角线，定位到两个字符串中的始末位置。空间复杂度是O(M*N) , 时间复杂度O( M(M+N) ) （M<=N），主要的时间负载度在于字符串扫描和对角线扫描。

 

[程序实现] 2-待续

 

1-  求一个字符串中存在的最长的重复子串[后缀数组]

 

/* * main.cpp * * Created on: 2009-7-12 * Author: NeeSky */#define MAXSIZE 100#include <iostream>#include <vector>using namespace std;string strGiven = ""; /*The Global String*/int IndexArray[MAXSIZE]; /*The Array Index Array*//*** * Just initilize the IndexArray */void init_strGivenAndIndexArray(string strgiven ){ strGiven = strgiven; for (unsigned int i = 0; i < strgiven.length(); ++i) IndexArray[i] = i; return;}/*** * Just for qsort args */inline int compare_OfIndexArray(const void *Index1, const void *Index2 ){ int* index1 = (int *) Index1; int* index2 = (int *) Index2; if ((string)&strGiven[*index1]> (string)&strGiven[*index2]) return 1; //string compare if ((string)&strGiven[*index1] == (string)&strGiven[*index2]) return 0; else return -1;}/*** * make the string to be order * @param strGiven */inline void enOrder_OfIndexArray(){ std::qsort(IndexArray, strGiven.length(), sizeof(int), compare_OfIndexArray); return;}int getCommonLength(string &comString, string a, string b ){ int i = 0; int k = a.length() < b.length() ? a.length() : b.length(); while (a[i] == b[i] && i < k) ++i; comString = a.substr(0, i); return i;}/*** * [Core Algorithm] Get the max repeat substring * @param strgiven */void getMaxRepeatSubString(string strgiven ){ vector<string> vecComStr; /*The same len substring, not only one*/ int maxlen = 0; init_strGivenAndIndexArray(strgiven); /*Initilize the global variables*/ enOrder_OfIndexArray(); /*enorder of array of postfix array of string*/ for (unsigned int i = 0; i < strgiven.length() - 1; ++i) /* Compare the neighbor*/ { string strTemp = ""; int templen = getCommonLength(strTemp, (string) &strGiven[IndexArray[i]], (string) &strGiven[IndexArray[i + 1]]); if (templen > maxlen) { maxlen = templen; vecComStr.clear(); vecComStr.push_back(strTemp); }else if(templen==maxlen) /*Max len substring not only one*/ vecComStr.push_back(strTemp); } cout << "The max length Repeat SubString of /"" << strgiven << "/"(maxlen is " << maxlen << ") : "; for(vector<string>::iterator it = vecComStr.begin() ; it != vecComStr.end(); ++it) /*for output*/ { if(it!=vecComStr.begin())cout <<" or "; cout <<"/""<<*it<<"/""; } cout<<endl; return;}/*** * The Main Programming * @return */int main(void ){ getMaxRepeatSubString("abcdabcdefghefgh"); getMaxRepeatSubString("abdddiedddie239000"); getMaxRepeatSubString("XXXXYYYYYYY"); getMaxRepeatSubString("XXXXXXX"); return 0;}

 

1-测试用例输出

The max length Repeat SubString of "abcdabcdefghefgh"(maxlen is 4) : "abcd" or "efgh"
The max length Repeat SubString of "abdddiedddie239000"(maxlen is 5) : "dddie"
The max length Repeat SubString of "XXXXYYYYYYY"(maxlen is 6) : "YYYYYY"
The max length Repeat SubString of "XXXXXXX"(maxlen is 6) : "XXXXXX"