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Solution.java
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42 lines (40 loc) · 1.74 KB
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package LeetCode.String.HuiWen.Solution2;
public class Solution {
public String longestPalindrome(String s) {
//判断所给字符串是否为空串或者只有一个字符,如果是直接返回该字符串
if(s.equals("")||s.length()==1)
return s;
//创建s.length()*s.length()]的二维数组
int[][] dp = new int[s.length()][s.length()];
//创建字符数组,并将所给字符串转为字符数组存储在该数组中
char[] num = s.toCharArray();
int start = 0;
int maxLen = 1;//最长回文串长度
//i表示正在访问的字符
for(int i=0;i<s.length();i++){
dp[i][i] = 1;//将二位数组初始化,所有元素都为1
//如果当前正在访问的字符不是字符串的最后一个字符,并且该字符和字符串的下一个字符相等
//即该子串回文串只有两个相等的字符
if(i+1<s.length() && num[i] == num[i+1]) {
dp[i][i+1] = 1;
start = i;
maxLen = 2;
}
}
// 必须先把相同字母的回文串去除,然后把len从2开始计算
for(int len=2;len<num.length;len++)
{
for(int i=0;i<num.length;i++)
{
if (i + len < num.length && num[i] == num[i + len]) {
dp[i][i + len] = dp[i + 1][i + len - 1];//验证dp[i][i+len]是回文串
if (dp[i][i + len] ==1 && len + 1 > maxLen) {//记录下最长回文串的起始位置和长度
maxLen = len + 1;
start = i;
}
}
}
}
return new String(num,start,maxLen);
}
}