Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.
求字符串中的最长回文字符串;
我的思路比较简单,但是实现起来代码看着好乱。。。从第一个字符开始遍历,直到最后一个字符,对每一个在index位置的字符进行判断;
如果最长回文字符串是奇数,则判断--index和++index是否相等;
如果最长回文字符串是偶数,则判断index和++index是否相等;
class Solution {
public:
string longestPalindrome(string s) {
int maxlen = 0, index = 0, length;
if (s.size() < 3){
return s;
}
for (int i = 0; i < s.size()-1; ++i){
length = findPalindrome(s, i);
if (maxlen < length){
maxlen = length;
index = i;
}
}
if (maxlen % 2 == 0)
index = index - maxlen/2 + 1;
else
index = index - maxlen/2;
string longestSubString(s, index, maxlen);
return longestSubString;
}
int findPalindrome(string &s, int index){
int l_index = index-1, r_index = index+1;
int length1 = 1, length2 = 0;
/*针对abvba这种奇数的情况*/
while((s[l_index--] == s[r_index++]) && (r_index <= s.size())){
length1 += 2;
}
/*针对abba这种偶数的情况*/
l_index = index;
r_index = index+1;
while((s[l_index--] == s[r_index++]) && (r_index <= s.size())){
length2 += 2;
}
return (length1 > length2) ? length1 : length2;
}
};
虽然代码看着恶习,但是运行效率还阔以,Runtime: 72 ms;我再找找其它的方法;