leetcode-algorithms-5 Longest Palindromic Substring
leetcode-algorithms-5 Longest Palindromic Substring
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd"
Output: "bb"
解法1
暴力破解,从最长的长度开始查找所有的子串,检查子串是否为回文串.
class Solution
{
public:
bool checkPalindrome(const std::string &s)
{
int len = s.size();
for (int i = 0; i < len/2; ++i)
{
if (s[i] != s[len -i - 1])
return false;
}
return true;
}
string longestPalindrome(string s)
{
std::string pal;
int len = s.size();
for (int i = len; i > 0; --i)
{
for (int start = 0; start + i <= len; ++start)
{
pal = s.substr(start, i);
if (checkPalindrome(pal))
return pal;
}
}
return pal;
}
};
时间复杂度: O(n^3).三个嵌套循环,O(n) * O(n) * O(n/2) = O(n^3).
空间复杂度: O(1)
解法2
观察回文串的规律.如:aba,先找到一个字母,然后两边都增加相同字母也是回文串,以此类推,就可以找到最大回文串.还有另种情况是:abba,则要先找到两个相邻是相同的字母.因此关键就是找到这个回文串的中心.
class Solution
{
public:
int palindromeCenter(const std::string &s, int left, int right)
{
while(left >= 0 && right < s.size() && s[left] == s[right])
{
--left;
++right;
}
return right - left - 1;
}
string longestPalindrome(string s) {
int len = s.size();
if (len <= 0) return "";
int start = 0;
int palindromeLen = 0;
for (int i = 0; i < len; ++i)
{
int len1 = palindromeCenter(s, i, i);
int len2 = palindromeCenter(s, i, i + 1);
int maxLen = len1 > len2 ? len1 : len2;
if (maxLen > palindromeLen)
{
start = i - (maxLen - 1) / 2;
palindromeLen = maxLen;
}
}
return s.substr(start,palindromeLen);
}
};
时间复杂度: O(n^2).
空间复杂度: O(1).
