字符串中连续出现最多的子串 & 字符串中最长反复子串

    字符串中连续出现最多的子串 & 字符串中最长反复子串，这两个问题都能够用后缀数组来表示，至于后缀数组能够參考编程珠玑P156；后缀数组就是定义一个数组指针，分别指向字符串中的相应位置，例如以下：

a b c a b c a b c d e .substr[0]

b c a b c a b c d e ....substr[1]

c a b c a b c d e .......substr[2]

a b c a b c d e ..........substr[3]

b c a b c d e .............substr[4]

c a b c d e ...............substr[5]

a b c d e .................substr[6]

b c d e ...................substr[7]

c d e .....................substr[8]

d e ........................substr[9]

e ..........................substr[10]

一、字符串中连续出现最多的子串

<span style="font-size:18px;">string MaxTimesOfContinue(string str)
{
	int len = str.length();
	int maxCount = 0;
	string longest = "";
	for (int i = 0; i < len; ++i)
	{
		for (int j = i + 1; j < len; ++j)
		{
			if (str.substr(i, j - i) == str.substr(j, j - i))
			{
				int offset = j - i;
				int count = 2;
				for (int k = j + offset; j <= len; k += offset)
				{
					if (str.substr(i, offset) == str.substr(k, offset))
						++count;
					else
						break;
				}
				if (count > maxCount)
				{
					maxCount = count;
					longest = str.substr(i, offset);
				}
			}
		}
	}
	return longest;
}</span>

   这个问题相同能够用后缀数组的思想来做，当然開始肯定想到的是暴力法，即求全部反复子串的长度，之后选择一个最长的就可以。

int Comlen(char *str1, char *str2)
{
	int i = 0;
	while(*str2 && (*str1++ == *str2++))
		++i;
	return i;
}

int MaxLength(char *str)
{
	if(str == NULL)
		return 0;
	int maxLen = 0;
	int n = strlen(str);
	int maxi, maxj;
	for (int i = 0; i < n; ++i)
	{
		for(int j = i + 1; j < n; ++j)
		{
			int thisLen = 0;
			if ((thisLen = Comlen(&str[i], &str[j])) > maxLen)
			{
				maxLen = thisLen;
				maxi = i;
				maxj = j;
			}
		}
	}
	return maxLen;
}

将后缀数组按字典排序

a[0]:a
a[1]:ana
a[2]:anana
a[3]:banana 
a[4]:na
a[5]:nana

之后比較相邻两个子串就可以：

int Comlen(char *str1, char *str2)
{
	int i = 0;
	while(*str2 && (*str1++ == *str2++))
		++i;
	return i;
}
int	Pstrcmp(const void *a, const void *b)
{
	return  strcmp(*(char**)a, *(char**)b); 
} 
//char *a[11];
int MaxLength(char *str)
{
	if(str == NULL)
		return 0;
	int maxLen = 0;
	int len = strlen(str);
	char **a = new char *[len + 1];
	for (int i = 0; i < len ; ++i)
		a[i] = &str[i];
	qsort(a, len , sizeof(char *), Pstrcmp);
	for (int i = 0; i < len - 1; ++i)
		if(Comlen(a[i], a[i+1]) > maxLen)
			maxLen = Comlen(a[i], a[i+1]);
	return maxLen;
}