Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.
解题思路一：
暴力枚举
共N^2量级个子串（从下标零开始），每次检查需一个for循环，等于是3重for循环，时间复杂度O(n^3)
解题思路二:
动态规划
设定一个表格table[][]，其中table[i][j]表示substring（i，j+1）是不是Palindromic，时间复杂度为O(n^2)空间复杂度也为O(n^2)。
Java代码如下：
static public String longestPalindrome(String s) {
if(s.length()==1)return s;
int[][] table=new int[s.length()][s.length()];
int beginIndex = 0,endIndex = 0;
//初始化第一、第二条斜线
for(int i=0;i<s.length();i++){
table[i][i]=1;
if(i==s.length()-1)break;
if(s.charAt(i)==s.charAt(i+1)){
table[i][i+1]=1;
beginIndex=i;
endIndex=i+1;
}
}
//给第k条斜线赋值
for(int k=2;k<s.length();k++){
for(int i=0;i<s.length()-k;i++){
if(table[i+1][i+k-1]==1&&s.charAt(i)==s.charAt(i+k)){
table[i][i+k]=1;
beginIndex=i;
endIndex=i+k;
}
}
}
printTable(table);
return s.substring(beginIndex,endIndex+1);
}
public static void printTable(int table[][]){
for(int i=0;i<table.length;i++){
for(int j=0;j<table[i].length;j++){
System.out.print(table[i][j]+" ");
}
System.out.println("");
}
}
提交结果，TimeExceed。证明时间复杂度为O(n^2)是不能提交通过的。
解题思路三：
中心法，对S中每一个字符及重复的双字符为中心，进行遍历。时间复杂度为O(n^2)，在leetcode上竟然Accepted！
Java代码如下：
static public String longestPalindrome(String s) {
if (s.length() == 1) return s;
String longest = s.substring(0, 1);
for (int i = 0; i < s.length(); i++) {
// 检查单字符中心
String tmp = helper(s, i, i);
if (tmp.length() > longest.length())
longest = tmp;
// 检查多字符中心
tmp = helper(s, i, i + 1);
if (tmp.length() > longest.length())
longest = tmp;
}
return longest;
}
public static String helper(String s, int begin, int end) {
while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
begin--;
end++;
}
return s.substring(begin + 1, end);
}
解题思路四：
Manacher’s algorithm，时间复杂度为O(n)
算法思路比较复杂，参考链接
http://blog.csdn.net/ggggiqnypgjg/article/details/6645824
http://blog.csdn.net/xingyeyongheng/article/details/9310555
Java代码
static public String longestPalindrome(String s) {
char[] sChar = new char[2 * s.length() + 1];
for (int i = 0; i < sChar.length; i++) {
if (i % 2 == 0)
sChar[i] = '#';
else
sChar[i] = s.charAt(i / 2);
}
int[] p = new int[2 * s.length() + 1];
int id = 0, mx = 0, maxID = 0;
for (int i = 0; i < sChar.length; i++) {
// 核心算法
if (mx > i) {
p[i] = Math.min(p[2 * id - i], mx - i);
} else
p[i] = 1;
int low = i - p[i], high = i + p[i];
while (low >= 0 && high <= (sChar.length - 1)) {
if (sChar[low] == sChar[high]) {
p[i]++;
low--;
high++;
} else
break;
}
// 更新id和mx的值
if (i + p[i] > mx) {
id = i;
mx = id + p[i];
}
// 更新取得最大p【i】的id
if (p[maxID] < p[i])
maxID = i;
}
char[] result = new char[p[maxID] - 1];
for (int i = 0; i < result.length; i++) {
result[i] = sChar[maxID - p[maxID] + 2 + 2 * i];
}
return new String(result);
}
C++实现如下：
1 #include<string> 2 #include<vector> 3 #include<algorithm> 4 using namespace std; 5 class Solution { 6 public: 7 string longestPalindrome(string s) { 8 vector<char> sChar(2 * s.length() + 1, '#'); 9 for (int i = 0; i < sChar.size(); i++) 10 if(i&1) 11 sChar[i]= s[i / 2]; 12 vector<int> p(2 * s.length() + 1,0); 13 int id = 0, mx = 0, maxID = 0; 14 for (int i = 0; i < sChar.size(); i++) { 15 // 核心算法 16 if (mx > i) { 17 int temp = p[2 * id - i]; 18 p[i] =min(temp, mx - i); 19 } 20 else 21 p[i] = 1; 22 int low = i - p[i], high = i + p[i]; 23 while (low >= 0 && high <= (sChar.size() - 1)) { 24 if (sChar[low] == sChar[high]) { 25 p[i]++; 26 low--; 27 high++; 28 } 29 else 30 break; 31 } 32 if (i + p[i] > mx) { 33 id = i; 34 mx = id + p[i]; 35 } 36 if (p[maxID] < p[i]) 37 maxID = i; 38 } 39 vector<char> result(p[maxID] - 1,'0'); 40 for (int i = 0; i < result.size(); i++) { 41 result[i] = sChar[maxID - p[maxID] + 2 + 2 * i]; 42 } 43 string res; 44 res.assign(result.begin(), result.end()); 45 return res; 46 } 47 };