Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.
Example:
Input: “cbbd”
Output: “bb”
方法一:
Brute Force,时间
,超时,空间
class Solution {
public:
string longestPalindrome(string s) {
if(s.empty() || s.size() == 1)
return s;
int len = s.size(), maxlen = 0, start=0;
for(int i = 0; i < len; i++ )//子串的长度
{
for(int j =0; j< len - i; j++)//子串起始的位置
{
if(isPalindrome(s,i,j) && (i+1)>maxlen)
{
maxlen = i+1;
start = j;
}
}
}
return s.substr(start,maxlen);
}
private:
bool isPalindrome(string s ,int len ,int start)
{
int left = start, right = start + len;
while(left < right)
{
if(s[left] == s[right])
{
left ++;
right --;
}
else
return false;
}
return true;
}
};
方法二:DP,时间
,空间
参考:http://www.cnblogs.com/grandyang/p/4464476.html
此题还可以用动态规划Dynamic Programming来解,根Palindrome Partitioning II 拆分回文串之二的解法很类似,我们维护一个二维数组dp,其中dp[i][j]表示字符串区间[i, j]是否为回文串,当i = j时,只有一个字符,肯定是回文串,如果i = j + 1,说明是相邻字符,此时需要判断s[i]是否等于s[j],如果i和j不相邻,即i - j >= 2时,除了判断s[i]和s[j]相等之外,dp[j + 1][i - 1]若为真,就是回文串,通过以上分析,可以写出递推式如下:
dp[i, j] = 1 if i == j
= s[i] == s[j] if j = i + 1
= s[i] == s[j] && dp[i + 1][j - 1] if j > i + 1
这里有个有趣的现象就是如果我把下面的代码中的二维数组由int改为vector
class Solution {
public:
string longestPalindrome(string s) {
if(s.size() < 2)
return s;
int len = s.size(), maxlen = 0, start=0 ,end = 0;
int dp[len][len] = {0};//区间[i,j]
for(int i = 0; i < len; ++i )
{
for(int j =0; j <= i; ++j)//子串起始的位置
{
dp[j][i] = (s[i] == s[j] && (i - j < 2 || dp[j + 1][i - 1]));
if(dp[j][i] && i - j + 1 > maxlen)
{
maxlen = i - j + 1;
start = j;
end = i;
}
}
//dp[i][i] = 1;
}
return s.substr(start, end-start+1);
}
};
方法三:中心扩散法,时间 ,空间
class Solution {
string res="";//设置全局变量
public:
string longestPalindrome(string s) {
if(s.size() < 2)
return s;
for(int i = 0; i < s.size(); ++i)
{
helper(s,i,i);
helper(s,i,i+1);//无中心点,偶数情况
}
return res;
}
private:
void helper(string s ,int left, int right)
{
while(left >= 0 && right < s.size() && s[left] == s[right])
{
left --;
right ++;//中心扩散
}//跳出时,left+1到right-1为回文,
string cur = s.substr(left+1, right-left-1);//len=right-1-(left+1)+1=right-left-1
if(cur.size() > res.size())
res = cur;
}
};