【JAVA、C++】LeetCode 005 Longest Palindromic Substring

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：

 #include<string>
 #include<vector>
 #include<algorithm>
 using namespace std;
 class Solution {
 public:
     string longestPalindrome(string s) {
         vector<char> sChar( * s.length() + , '#');
         for (int i = ; i < sChar.size(); i++)
             if(i&)
                 sChar[i]= s[i / ];
         vector<int> p( * s.length() + ,);
         int id = , mx = , maxID = ;
         for (int i = ; i < sChar.size(); i++) {
             // 核心算法
             if (mx > i) {
                 int temp = p[ * id - i];
                 p[i] =min(temp, mx - i);
             }
             else
                 p[i] = ;
             int low = i - p[i], high = i + p[i];
             while (low >=  && high <= (sChar.size() - )) {
                 if (sChar[low] == sChar[high]) {
                     p[i]++;
                     low--;
                     high++;
                 }
                 else
                     break;
             }
             if (i + p[i] > mx) {
                 id = i;
                 mx = id + p[i];
             }
             if (p[maxID] < p[i])
                 maxID = i;
         }
         vector<char> result(p[maxID] - ,'');
         for (int i = ; i < result.size(); i++) {
             result[i] = sChar[maxID - p[maxID] +  +  * i];
         }
         string res;
         res.assign(result.begin(), result.end());
         return res;
     }
 };