5. Longest Palindromic Substring LeetCode题解

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

题意：

给定一个字符串s，找出其中最长的回文子串，假定字符串最长为1000

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

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题解：



详见：http://blog.csdn.net/baidu_23318869/article/details/48154899
Code【Java】

public class Solution {
// 在原字符串中插入指定字符
public String encode(String s, String ch) {
StringBuilder ss = new StringBuilder();
for (int i = 0; i < s.length(); ++i) {
ss.append(ch);
ss.append(s.charAt(i));
}
ss.append(ch);
return ss.toString();
}
// 删除原字符串中指定字符
public String decode(String s, String ch) {
return s.replaceAll( ch, "");
}
// 返回以当前pos为中心回文长度的一半（不含pos）
int getHalfPalindromeLen(String s, int pos) {
int i = 1;
for ( ; 0 <= pos - i && pos + i < s.length(); ++i) {
if (s.charAt(pos - i) != s.charAt(pos + i)) {
break;
}
}
return i - 1;
}

public String longestPalindrome(String s) {
// 初始化字符串以及记录回文长度的数组
s = encode(s, "#");
int[] halfLen = new int[s.length()];
// 上一次最长回文串位置
int lastPos = 0;
int lastLen = 0;
for (int i = 0; i < s.length(); ++i) {
if (i <= lastPos + lastLen) {
int dis = i - lastPos;
int cor = lastPos - dis;
halfLen[i] = (cor - halfLen[cor] <= lastPos - lastLen) ?
getHalfPalindromeLen(s, i) : halfLen[cor];
}
else {
halfLen[i] = getHalfPalindromeLen(s, i);
}
if (lastPos + lastLen < i + halfLen[i]) {
lastPos = i;
lastLen = halfLen[i];
}
}
// 寻找最长回文串并返回
int maxPos = 0;
for (int i = 1; i < halfLen.length; ++i) {
if (halfLen[maxPos] < halfLen[i]) {
maxPos = i;
}
}
return decode(s.substring(maxPos - halfLen[maxPos], maxPos + halfLen[maxPos] + 1), "#");
}

}

Code【C++】

class Solution {
// 在原字符串中插入指定字符
string encode(string s, char ch = '#') {
stringstream ss;
for (int i = 0; i < s.size(); ++i) {
ss << ch << s[i];
}
ss << ch;
return ss.str();
}
// 删除原字符串中指定字符
string decode(string s, char ch = '#') {
stringstream ss;
for (int i = 0; i < s.size(); ++i) {
if (s[i] != ch) {
ss << s[i];
}
}
return ss.str();
}
// 返回以当前pos为中心回文长度的一半（不含pos）
int getHalfPalindromeLen(string s, int pos) {
int i = 1;
for ( ; 0 <= pos - i && pos + i < s.size(); ++i) {
if (s[pos - i] != s[pos + i]) {
break;
}
}
return i - 1;
}
public:
string longestPalindrome(string s) {
// 初始化字符串 以及记录最长回文长度数组
s = encode(s);
vector<int> halfLen(s.size(), 0);

// 开始计算最长回文
int lastPos = 0;
int lastLen = 0;
for (int i = 0; i < s.size(); ++i) {
if (i <= lastPos + lastLen) {
int dis = i - lastPos;
int cor = lastPos - dis;
halfLen[i] = (cor - halfLen[cor] <= lastPos - lastLen) ?
getHalfPalindromeLen(s, i) : halfLen[cor];
}
else {
halfLen[i] = getHalfPalindromeLen(s, i);
}
// 更新lastPos以及lastLen
if (lastPos + lastLen < i + halfLen[i]) {
lastPos = i;
lastLen = halfLen[i];
}
}
// 寻找最长回文位置
int maxPos = 0;
for (int i = 1; i < s.size(); ++i) {
if (halfLen[maxPos] < halfLen[i]) {
maxPos = i;
}
}
return decode(s.substr(maxPos - halfLen[maxPos], halfLen[maxPos] * 2 + 1));
}
};