最长回文子串
题目 中等
和最长回文子序列类似
自己的做法:
class Solution {
public String longestPalindrome(String s) {
int len = s.length();
int max = 1;
int left = 0, right = 0;
int[][] dp = new int[len][len];
dp[0][0] = 1;
for (int j = 1; j < len; j++) {
dp[j][j] = 1;
for (int i = j - 1; i >= 0; i--) {
if (s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1] == j - 1 - i) {
dp[i][j] = dp[i + 1][j - 1] + 2;
if (dp[i][j] > max) {
max = dp[i][j];
left = i;
right = j;
}
}
else
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
}
}
return s.substring(left, right + 1);
}
}
官方解法:
public class Solution {
public String longestPalindrome(String s) {
// 特殊用例判断
int len = s.length();
if (len < 2) {
return s;
}
int maxLen = 1;
int begin = 0;
// dp[i][j] 表示 s[i, j] 是否是回文串
boolean[][] dp = new boolean[len][len];
char[] charArray = s.toCharArray();
for (int i = 0; i < len; i++) {
dp[i][i] = true;
}
for (int j = 1; j < len; j++) {
for (int i = 0; i < j; i++) {
if (charArray[i] != charArray[j]) {
dp[i][j] = false;
} else {
if (j - i < 3) {
dp[i][j] = true;
} else {
dp[i][j] = dp[i + 1][j - 1];
}
}
// 只要 dp[i][j] == true 成立,就表示子串 s[i..j] 是回文,此时记录回文长度和起始位置
if (dp[i][j] && j - i + 1 > maxLen) {
maxLen = j - i + 1;
begin = i;
}
}
}
return s.substring(begin, begin + maxLen);
}
}
作者:liweiwei1419
链接:https://leetcode.cn/problems/longest-palindromic-substring/solutions/7792/zhong-xin-kuo-san-dong-tai-gui-hua-by-liweiwei1419/
来源:力扣(LeetCode)
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标签:String,int,做过,len,maxLen,HOT100,除去,dp,回文 From: https://www.cnblogs.com/CWZhou/p/17537245.html