給定兩個字符串，求它們前後重疊的最長子串的長度，比如"abcde"和“cdefg”是"cde"，長度爲3。

#include <iostream>
#include <string>
using namespace std;
 
//計算字符串的前綴函數
int* computePrefix(const char *p, int n)
{
	if(p == NULL || n <= 0) return NULL;
	int *pr = new int[n];
	memset(pr, -1, sizeof(int)*n);
	int k = -1;
	for(int i  = 1; i < n; i++)
	{
		while(k > -1 && p[k+1] != p[i]) k = pr[k];
		if(p[k+1] == p[i]) k = k+1;
		pr[i] = k;
	}//for
	return pr;
}//computePrefix
 
//返回兩個字符串的前後重疊的最長子串的長度
int LCS(const char* p, int n, const char* s, int m)
{
	if(p == NULL || n <= 0 || s == NULL || m <= 0) return 0;
	int *pr = computePrefix(s, m);
	int k = -1;
	for(int i = 0; i < n; i++)
	{
		while(k > -1 && s[k+1] != p[i]) k = pr[k];
		if(s[k+1] == p[i]) k = k+1;
 
		if(i != n-1 && k == m-1) k = pr[m-1];
	}
	delete[] pr;
	return k+1;
}//LCS
 
int main()
{
	string p, s;
	while(cin >> p >> s)
	{
		int l1 = LCS(p.c_str(), p.size(), s.c_str(), s.size());
		int l2 = LCS(s.c_str(), s.size(), p.c_str(), p.size());
		int lcs = (l1>l2)?l1:l2;
		cout << lcs << endl;
	}
 
	getchar();
}