算法一：O(n^3)
判断字串是否对称是从外到里， O(n)
/*
*判断起始指针，到结束指针的字符串是否对称
*/
int IsSymmetrical(char* pBegin, char* pEnd)
{
if(pBegin == NULL || pEnd == NULL || pBegin > pEnd)
return 0;
while(pBegin < pEnd)
{
if(*pBegin != *pEnd)
return 0;
pBegin++;
pEnd--;
}
return 1;
}
/*
*查找最大对称字串长度，时间复杂度是O(n^3)
*/
int GetLongestSymmetricalLength(char* pString)
{
if(pString == NULL)
return 0;
int symmetricalLength = 1;
char* pFirst = pString;
int length = strlen(pString);
while(pFirst < &pString[length-1])
{
char* pLast = pFirst + 1;
while(pLast <= &pString[length-1])
{
if(IsSymmetrical(pFirst, pLast))
{
int newLength = pLast - pFirst + 1;
if(newLength > symmetricalLength)
symmetricalLength = newLength;
}
pLast++;
}
pFirst++;
}
return symmetricalLength;
}
int main()
{
char* str = "google";
int len = GetLongestSymmetricalLength(str);
printf("%d", len);
return 0;
}
判断字串是否对称是从外到里， O(1)
int GetLongestSymmetricalLength(char* pString)
{
if(pString == NULL)
return 0;
int symmetricalLength = 1;
char* pChar = pString;
while(*pChar != '\0')
{
//奇数长度对称， 如 aAa
char* left = pChar - 1;
char* right = pChar + 1;
while(left >= pString && *right != '\0' && *left==*right)
{
left--;
right++;
}
int newLength = right - left - 1; //退出循环是*left!=*right
if(newLength > symmetricalLength)
symmetricalLength = newLength;
//偶数长度对称， 如 aAAa
left = pChar;
right = pChar + 1;
while(left >= pString && *right != '\0' && *left==*right)
{
left--;
right++;
}
newLength = right - left - 1; //退出循环是*left!=*right
if(newLength > symmetricalLength)
symmetricalLength = newLength;
pChar++;
}
return symmetricalLength;
}
int main()
{
char* str = "google";
int len = GetLongestSymmetricalLength(str);
printf("%d", len);
return 0;
}
算法3：manacher算法
原串：abaab
新串：#a#b#a#a#b#
这样一来，原来的奇数长度回文串还是奇数长度，偶数长度的也变成以‘#'为中心的奇数回文串了。
接下来就是算法的中心思想，用一个辅助数组P记录以每个字符为中心的最长回文半径，也就是P[i]记录以Str[i]字符为中心的最长回文串半径。P[i]最小为1，此时回文串为Str[i]本身。
我们可以对上述例子写出其P数组，如下
新串： # a # b # a # a # b #
P[] : 1 2 1 4 1 2 5 2 1 2 1
我们可以证明P[i]-1就是以Str[i]为中心的回文串在原串当中的长度。
证明：
1、显然L=2*P[i]-1即为新串中以Str[i]为中心最长回文串长度。
2、以Str[i]为中心的回文串一定是以#开头和结尾的，例如“#b#b#”或“#b#a#b#”所以L减去最前或者最后的‘#'字符就是原串中长度的二倍，即原串长度为(L-1)/2，化简的P[i]-1。得证。
注: 不是很懂， 自己改了
int GetLongestSymmetricalLength(char* pString)
{
int length = strlen(pString);
char* pNewString = malloc(2*length+2);
int i;
for(i=0; i<length; i++)
{
*(pNewString+i*2) = '#';
*(pNewString+i*2+1) = *(pString+i);
}
*(pNewString+2*length) = '#';
*(pNewString+2*length+1) = '\0';
printf("%s\n", pNewString);
int maxLength = 1;
char* pChar;
for(i=0; i<2*length+2; i++)
{
int newLength = 1;
pChar = pNewString + i;
char* left = pChar-1;
char* right = pChar+1;
while(left>=pNewString && *right!='\0'&& *left==*right)
{
left--;
right++;
newLength++;
}
if(newLength > maxLength)
maxLength = newLength;
}
return maxLength-1;
}
int main()
{
char* str = "google";
int len = GetLongestSymmetricalLength(str);
printf("%d", len);
return 0;
}
