[LeetCode OJ]Longest Palindromic Substring
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【问题描述】
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.
Example:
Input: "cbbd"Output: "bb"问题来源:longest-palindromic-substring
【解题分析】
要寻找一个最长的连续回文子串。
我们定义dp[i][j]为从i到j的字符串是不是回文串。
依据回文的规则:
1. 如果s[i] = s[j],那么是否是回文串决定于dp[i+1][j-1]
2. 如果s[i]!=s[j],那么dp[i][j] = 0
3. 动态规划是按照字符串的长度从1到n进行(不然dp初始化为0,就算整个字符串是一个回文串,它也只会截取到某一子回文串)。
代码时间复杂度为O(N^2),空间复杂度为O(N^2)。
我们定义dp[i][j]为从i到j的字符串是不是回文串。
依据回文的规则:
1. 如果s[i] = s[j],那么是否是回文串决定于dp[i+1][j-1]
2. 如果s[i]!=s[j],那么dp[i][j] = 0
3. 动态规划是按照字符串的长度从1到n进行(不然dp初始化为0,就算整个字符串是一个回文串,它也只会截取到某一子回文串)。
代码时间复杂度为O(N^2),空间复杂度为O(N^2)。
【源代码】
class Solution {public: string longestPalindrome(string s) { vector<vector<bool> > dp(s.length(),vector<bool>(s.length(),0)); if (s.length() == 0 || s.length() == 1) { return s; } for (int i = 0; i < s.length(); i++) { for (int j = 0; j < s.length(); j++) { if (i >= j) { dp[i][j] = 1; } } } int maxLength = 1; int left = 0; int right = 0; for (int k = 1; k < s.length(); k++) { for (int i = 0; k + i < s.length(); i++) { int j = k + i; if (s[i] != s[j]) { dp[i][j] = 0; } else { dp[i][j] = dp[i+1][j-1]; if (dp[i][j]) { if (k + 1 > maxLength) { maxLength = k + 1; left = i; right = j; } } } } } return s.substr(left,right-left+1); }};
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