我的做法就是暴力，再加点剪枝来不超时。具体的思路是先从整体去识别这个字符串是否为回文串，如果不是再用连续的n-1个字符进行判断是否为回文，一旦是就不用继续判断其他的了，已经是最长的回文了，所以可以减少很多判断，效率比较高。
public String judge(String str) {
int commonLength;
String result = str.substring(0, 1);
if (str.length() % 2 == 0) {
commonLength = 0;
int index = str.length() / 2;
int left = index - 1;
int right = index;
while (true) {
if (left < 0 || right >= str.length()) {
break;
}
if (str.charAt(left) == str.charAt(right)) {
commonLength += 2;
result = str.substring(left, right + 1);
left--;
right += 1;
} else {
break;
}
}
} else {
commonLength = 1;
int index = str.length() / 2;
int left = index - 1;
int right = index + 1;
while (true) {
if (left < 0 || right >= str.length()) {
break;
}
if (str.charAt(left) == str.charAt(right)) {
result = str.substring(left, right + 1);
commonLength += 2;
left--;
right += 1;
} else {
break;
}
}
}
return result;
}
public String longestPalindrome(String s) {
String commonStr = s.substring(0, 1);
int commonLength = s.length();
int resultLength = 1;
while (true) {
for (int i = 0; i < s.length(); i++) {
if (commonLength + i <= s.length()) {
String result = judge(s.substring(i, i + commonLength));
if (result.length() > commonStr.length()) {
commonStr = result;
resultLength = commonStr.length();
}
} else {
break;
}
}
commonLength--;
if (commonLength <= resultLength) break;
}
return commonStr;
}中心扩散法
思路大同小异，这个是从小扩大，我的是从大扩小，我的要高效些。
public String longestPalindrome1(String s) {
if (s == null || s.length() == 0) {
return "";
}
int strLen = s.length();
int left = 0;
int right = 0;
int len = 1;
int maxStart = 0;
int maxLen = 0;
for (int i = 0; i < strLen; i++) {
left = i - 1;
right = i + 1;
while (left >= 0 && s.charAt(left) == s.charAt(i)) {
len++;
left--;
}
while (right < strLen && s.charAt(right) == s.charAt(i)) {
len++;
right++;
}
while (left >= 0 && right < strLen && s.charAt(right) == s.charAt(left)) {
len = len + 2;
left--;
right++;
}
if (len > maxLen) {
maxLen = len;
maxStart = left;
}
len = 1;
}
return s.substring(maxStart + 1, maxStart + maxLen + 1);
}
作者：reedfan
链接：https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zhong-xin-kuo-san-fa-he-dong-tai-gui-hua-by-reedfa/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。动态规划。
记录已经计算过的区间的是否为回文。
public String longestPalindrome(String s) {
if (s == null || s.length() < 2) {
return s;
}
int strLen = s.length();
int maxStart = 0; //最长回文串的起点
int maxEnd = 0; //最长回文串的终点
int maxLen = 1; //最长回文串的长度
boolean[][] dp = new boolean[strLen][strLen];
for (int r = 1; r < strLen; r++) {
for (int l = 0; l < r; l++) {
if (s.charAt(l) == s.charAt(r) && (r - l <= 2 || dp[l + 1][r - 1])) {
dp[l][r] = true;
if (r - l + 1 > maxLen) {
maxLen = r - l + 1;
maxStart = l;
maxEnd = r;
}
}
}
}
return s.substring(maxStart, maxEnd + 1);
}
作者：reedfan
链接：https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zhong-xin-kuo-san-fa-he-dong-tai-gui-hua-by-reedfa/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
作者：你的雷哥
本文版权归作者所有，欢迎转载，但未经作者同意必须在文章页面给出原文连接，否则保留追究法律责任的权利。
