C++:求两个字符串的最大公有子串和字符串逆序

#include "iostream"
using namespace std;


#include<cassert>
#include <string>
//找出两个字符串中最长的公共子串（如果存在多个 ， 返回第一个）
//参数：str1　　 字符串1
//　　　str2     字符串2
//　　　maxSubStr 　　 找到的最大子串
void findMaxSubstr(const char * str1, const char * str2, char * maxSubstr){
	assert((str1 != NULL) && (str2 != NULL));
	assert(maxSubstr != NULL);
	int maxPos = -1;
	int maxLen = 0,k;
	for (int i = 0; i<strlen(str1); i++){
		for (int j = 0; j<strlen(str2); j++){
			if (str1[i] == str2[j]){
				for (k = 1; (str1[i + k] == str2[j + k]) && (str1[i + k] != '\0'); k++);
				if (k>maxLen){
					maxPos = i;
					maxLen = k;
				}
			}
		}
	}
	if (maxPos == -1){
		maxSubstr[0] = '\0';
	}
	else{
		memcpy(maxSubstr, str1 + maxPos, maxLen);
		maxSubstr[maxLen] = '\0';
	}
}

//方法2
char* lcs(char *str1, char *str2)
{
	int len1, len2;
	int i, j;
	char *shorter, *longer, *substr;
	if (str1 == NULL || str2 == NULL)return NULL;

	len1 = strlen(str1);
	len2 = strlen(str2);

	if (len1 < len2)
	{
		shorter = str1;
		longer = str2;
	}
	else
	{
		shorter = str2;
		longer = str1;
	}

	if (strstr(longer, shorter) != NULL) return shorter;

	len1 = strlen(shorter);
	substr = (char*)malloc(len1 + 1);

	for (i = len1 - 1; i > 0; i--)
	{
		for (j = 0; j <= len1 - i; j++)
		{
			memcpy(substr, &shorter[j], i);
			substr[i] = '\0';
			if (strstr(longer, substr) != NULL) return substr;

		}
	}

	return NULL;


}


//方法3
string lcs_search(string str1, string str2)//求两个字符串的公共子串  
{
	int len1 = str1.length();
	int len2 = str2.length();
	int length = 0;
	int end = 0;
	if (len1 < len2)
	{
		string temp = str1;
		str1 = str2;
		str2 = temp;
		len1 = str1.length();
		len2 = str2.length();
	}
	int *sign = new int[len1];
	for (int i = 0; i < len2; i++)
	{
		for (int j = len1 - 1; j >= 0; j--)
		{
			if (str2[i] == str1[j])
			{
				if (i == 0 || j == 0)sign[j] = 1;
				else sign[j] = sign[j - 1] + 1;
			}
			else sign[j] = 0;
			if (sign[j] > length)
			{
				length = sign[j];
				end = j;
			}
		}
	}
	delete[]sign;
	return str1.substr(end - length + 1, length);

}


//方法4
string   FindMAxSubString(string a, string b)
{
	string res = "";
	string shorter = a.length()>b.length() ? b : a;
	string longer = a.length()>b.length() ? a : b;

	int maxlen = 0;
	int maxpos = -1;
	for (int i = 0; i<shorter.length(); i++){
		for (int j = 0; j<longer.length(); j++)
		{
			if (shorter[i] == longer[j])
			{
				int k = 1;
				for (; (shorter[i + k] == longer[j + k]) && i + k<shorter.length(); k++);
				if (k>maxlen){
					maxpos = i;
					maxlen = k;
				}
			}
		}
	}

	if (maxpos == -1)
		return res;
	else
	{
		res = shorter.substr(maxpos, maxlen);
		return res;
	}

}

int main(){

	char substr[20];

	findMaxSubstr("zhangligu", "gligzhangligu", substr);
	//cout << lcs("zhangligu", "gligzhangligu");
	//cout << lcs_search("zhangligu", "gligzhangligu");
	cout << FindMAxSubString("zhangligu", "gligzhangligu");
	
	//cout << substr << endl;

	return 0;

}

字符串逆序

char* str_reverse(char* str)
{
	int n = strlen(str) / 2;
	int i = 0;
	char tmp = 0;

	for (i = 0; i < n; i++)
	{
		tmp = str[i];
		str[i] = str[strlen(str) - i - 1];
		str[strlen(str) - i - 1] = tmp;
	}

	return str;
}

部分内容参考自：点击打开链接1点击打开链接2点击打开链接3