leetcode problem (5)

最长回文子串:

1. 暴力搜索   时间复杂度O(n^3)

2. 动态规划

  • dp[i][j] 表示子串s[i…j]是否是回文
  • 初始化:dp[i][i] = true (0 <= i <= n-1);  dp[i][i-1] = true (1 <= i <= n-1); 其余的初始化为false
  • dp[i][j] = (s[i] == s[j] && dp[i+1][j-1] == true)

  时间复杂度O(n^2),空间O(n^2)

3.  以某个元素为中心,分别计算偶数长度的回文最大长度和奇数长度的回文最大长度。时间复杂度O(n^2),空间O(1)

4. Manacher算法,时间复杂度O(n), 空间复杂度O(n)。 具体参考如下链接:

    http://articles.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html

  

class Solution {
public:
    string longestPalindrome(string s) {
        int n = 2*s.length() + 3;
        char cstr[n];
        cstr[0] = ‘\1‘;
        cstr[n-1] = ‘\0‘;
        for(int i = 1; i < n; i += 2)
            cstr[i] = ‘#‘;
        for(int i = 2; i < n; i += 2)
            cstr[i] = s[i/2 - 1];
        int *p;
        p = new int[n];
        memset(p, 0, sizeof(int)*n);

        int mx, id, i, j;
        for (id = 1, i = 2, mx = 0; i < n; ++i) {
            j = 2*id - i;
            p[i] = (mx > i) ? min(p[j], mx-i) : 0;
            while (cstr[i + p[i] + 1] == cstr[i - (p[i] + 1)])
                ++p[i];
            if (i + p[i] > mx){
                id = i;
                mx = id + p[id];
            }
        }

        int max = -1;
        for (i = 2; i < n-2; ++i) {
            if (max < p[i]) {
                max = p[i];
                id = i;
            }
        }
        return s.substr( (id - max - 1)/ 2 , max);
    }
private:

};

  

时间: 2024-10-15 15:00:36

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