Discription:
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”
分析
求一个字符串的最长回文子串,子串必须得是连续的,不是子序列。暴力方法就不想了,处理回文串的一种方法是dp动态规划。
dp[i][j]记录给定字符串i-j是否是回文子串,若是记为1,;否则为0
则:
dp[i][j]=(s[i]==s[j]&&dp[i+1][j-1])
也就是将全部dp先初始化为0,然后所有的长度为1的子串肯定为1.也就是dp[i][i]=1;
然后查长度为2的子串:若s[i]==s[i+1]则dp[i][i+1]=1;
然后遍历三个及以上长度的子串,开始为i,末端索引j=i+k-1 (k为子串长度)
用st,en分别记录子串开始和尾端索引,输出即可
代码
string longestPalindrome(string s) {
int len = s.size();
if (len <= 1)return s;
int st=0, en=0,l;//l记录最长回文子串长度,本题用不着
bool dp[1001][1001];
memset(dp, 0, sizeof(dp));
for (int i = 0; i<len; i++)
{
dp[i][i] = 1;
}
for (int i = 0; i<len - 1; i++)
{
if (s[i] == s[i + 1])
{
dp[i][i + 1] = 1;
l = 2;
st = i; en = i + 1;
}
}
for (int k = 3; k <= len; k++)
{
for (int i = 0; i + k - 1<len; i++){
int j = i + k - 1;//右端索引
if (s[j] == s[i] && dp[i + 1][j - 1])
{
dp[i][j] = 1;
l = k;
st = i; en = j;
}
}
}
string ans = "";
for (int k = st; k <= en; k++)
{
ans += s[k];
}
return ans;
}
注意
st en一开始就要初始化,否则可能在最后遍历st-en时运行时错误