Leetcode:5. Longest Palindromic Substring
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Description:
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”
解题思路:
本题为经典的最长回文子字符串问题,题意很清楚,就是求字符串S中的最长回文。一种比较好想到的算法思路就是中心扩展法,从下标 i 出发,用2个指针向 i 的两边扩展判断是否相等,那么只需要对0到n-1的下标都做此操作,就可以求出最长的回文子串。但需要注意的是,回文字符串有奇偶对称之分,即”abcba”与”abba”2种类型。
string longestPalindrome(string s) { int 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; //将max的高七位变成0,得到max的最低位 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; }
动态规划求解
假设dp[ i ][ j ]的值为true,表示字符串s中下标从 i 到 j 的字符组成的子串是回文串。那么可以推出: dp[ i ][ j ] = dp[ i + 1][ j - 1] && s[ i ] == s[ j ]。这是一般的情况,由于需要依靠i+1, j -1,所以有可能 i + 1 = j -1, i +1 = (j - 1) -1,因此需要求出基准情况才能套用以上的公式:
a. i + 1 = j -1,即回文长度为1时,dp[ i ][ i ] = true;
b. i +1 = (j - 1) -1,即回文长度为2时,dp[ i ][ i + 1] = (s[ i ] == s[ i + 1])。
需要注意的是动态规划需要额外的O(
class Solution {public: string longestPalindrome(string s) { if (s.length() == 0) return ""; if (s.length() == 1) return s; int len = s.length(); bool dp[len][len]; int start = 0, max = 1; for (int i = 0; i < len; i++) { dp[i][i] = true; if (i +1 < len) if (s[i] == s[i+1]) { dp[i][i+1] = true; start = i; max = 2; } else { dp[i][i+1] = false; } } //动态规划求解 for (int k = 3; k <= len; k++) { for (int i = 0; i < len-k+1; i++) { int j = i + k-1; if (dp[i+1][j-1] && s[i] == s[j]) { dp[i][j] = true; int cur = j-i+1; if (cur > max) { start = i; max = cur; } } else { dp[i][j] = false; } } } return s.substr(start,max); } };
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