LeetCode:Longest Palindromic Substring
2017-12-31 22:16
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题目:
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.Example:
Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.
注意,两个相等的字符是回文,单个字符本身也算回文。
思路:
建立一个数据类型为bool的二维数组table,初始化为全false。其中table[i][j]为true表示字符串s的下标i到j的部分为回文,false则表示不是回文。
动态规划类似于分治算法,要求原问题的解,先求出更小一些的子问题的解,由子问题的解通过某种方式得到原问题的解。
在这里我们要找字符串s的最长回文子字符串(LPS),那么可以先求出s减去首或尾一个字符的部分的LPS,进而可以再往前推,再减去首或尾的一个字符,然后求LPS……
因此,推到最后,我们会发现可以把原问题拆成s.size()个单字符,先遍历整个s求单个字符是不是回文(明显是),然后再遍历整个s求两个字符一组的子字符串是不是回文,然后再遍历整个s去判断三个字符一组的版本……直到最后遍历整个s去判断s.size()个字符一组的版本。
代码:
class Solution { public: string longestPalindrome(string s) { int size = s.size(); int begin = 0, maxlen = 1; bool table[1000][1000] = {false}; //单个字符的判断,明显得全是true for (int i = 0; i < size; i++) { table[i][i] = true; } //两个字符一组 for (int i = 0; i < size-1; i++) { if (s[i] == s[i+1]) { table[i][i+1] = true; begin = i; maxlen = 2; } } //三个字符及以上的一组遍历 for (int N = 3; N <= size; N++) { //拆len个字符出来作为一组 for (int a = 0; a < size-N+1; a++)//拆出来的子串首部a { int b = a+N-1;//子串尾部b if (s[a] == s[b] && table[a+1][b-1]) { table[a][b] = true; begin = a; maxlen = N; } } } return s.substr(begin, maxlen); } };
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