[LeetCode]Longest Palindromic Substring

題目要求如下：

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.

也就是傳說中的求最長迴文子串的問題，具體的算法分析請點本博客的文章《求最長迴文子串(Longest Palindromic Substring)》

嘗試用動態規劃和中心擴展法都做了一下，發現動態規劃的時間要比中心擴展法多出4倍左右，可能是因爲對數組的訪問導致了時間的增加，而中心擴展法只需要遍歷字串而不需要同時訪問2維數組，所以更快。另外在LeetCode中用動態數組會出錯，不知道什麼原因。

以下是2種方法的代碼，歡迎大牛指導交流~

1.動態規劃法

AC，Runtime: 552 ms

//LeetCode_Longest Palindromic Substring
//Written by zhou
//2013.11.22

class Solution {
public:
     string longestPalindrome(string s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.

        size_t n = s.length();
        /*bool **dp = new bool*[n];
        for (size_t i = 0; i < n; ++i)
        {
            dp[i] = new bool[n];
        }*/
        
        bool dp[1000][1000];
        
        //爲基準情況賦值
        int startPos = 0;
        int max = 1;
        for (size_t i = 0; i < n; ++i)
        {
            dp[i][i] = true;
            if (i + 1 < n)
            {
				if (s[i] == s[i+1])
				{
					dp[i][i+1] = true;
					startPos = i;
					max = 2;
				}
				else dp[i][i+1] = false;
            }
        }
        
        //動規求解,前面已求len = 1, len = 2的情況
        for (int len = 3; len <= n; ++len)
        {
            for (int i = 0; i < n - len + 1; ++i)
            {
                int j = i + len - 1;
                
				if (dp[i+1][j-1] && s[i] == s[j])
				{
					dp[i][j] = true;
					int curLen = j - i + 1;
					if (curLen > max)
					{
						startPos = i;
						max = curLen;
					}
				}
				else dp[i][j] = false;
                
            }
        }
        
		//釋放二維數組
        /*for (size_t i = 0; i < n; ++i)
           delete[] dp[i];
        
        delete[] dp;*/
        
		return s.substr(startPos,max);
    }
};

 

2. 中心擴展法

AC，Runtime: 132 ms

//LeetCode_Longest Palindromic Substring
//Written by zhou
//2013.11.22

class Solution {
public:
     string longestPalindrome(string s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
       
        size_t n = s.length();
        int startPos = 0;
        int max = 1;
        for (int i = 0; i < n; ++i)
        {
            int oddLen = 0, evenLen = 0, curLen;
            oddLen = Palindromic(s,i,i);
            
            if (i + 1 < n)
               evenLen = Palindromic(s,i,i+1);
            
            curLen = oddLen > evenLen? oddLen : evenLen;
            
            if (curLen > max)
            {
                max = curLen;
                if (max & 0x1)
                  startPos = i - max / 2;
                else 
                  startPos = i - (max - 1) / 2;
            }
        }
        
        return s.substr(startPos,max);
    }
    
    int Palindromic(const string &str, int i, int j)
    {
        size_t n = str.length();
        int curLen = 0;

        while (i >= 0 && j < n && str[i] == str[j])
        {
			--i;
            ++j;
        }
        curLen = (j-1) - (i+1) + 1;

        return curLen;
    }
};