【leetcode】5. Longest Palindromic Substring(Python & C++)
来源:互联网 发布:生化危机游戏 知乎 编辑:程序博客网 时间:2024/06/05 03:28
5. Longest Palindromic Substring
题目链接
5.1 题目描述:
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”
5.2 解题思路:
思路一:首先创建判断是否是回文的函数ishuiwen。需要三个参数,string s,int x,int y。x和y是判断字符串s的起始位置。如果x==y,则为奇数回文字符串判断,并且回文字符串起始长度设为1,x–,y++。如果x!=y,则为偶数回文字符串判断,并且回文字符串起始长度设为0。然后循环条件,x大于等于0 并且 y小于s.length() && s[x]==s[y],则长度+=2,x–,y++。这样最后返回得到的回文字符串s.substr(x + 1, count)。然后在longestPalindrome函数中,寻找最大的回文子串。从k=0开始,每次判断两个回文子串,一个奇数的,一个偶数的,并记录长度大的子串。
思路二:首先,设置回文子串起始位置first=0,长度count=1。for k循环遍历字符串s,首先设置临时长度c_temp=0。用i=k,然后for j逆向找个s中第一个与是s[i]相等的字符,这是回文子串的结束位置。此时回文子串的临时长度c_temp=2,然后i++,j–。如果i小于j,则继续判断s[i]==s[j],相等则i++,j–,c_temp+=2,不相等则c_temp=0,同时break。如果i==j,则c_temp+=1.然后判断c_temp与count的大小,决定是否更新count与first。这样直至循环结束。最后,返回s.substr(first,count)。
5.3 C++代码:
1、思路一代码(32ms):
class Solution {public: string ishuiwen(string s,int x,int y) { int count; if (x == y) { count = 1; x--; y++; } else count = 0; while (x>=0 && y<s.length() && s[x]==s[y] ) { x--; y++; count += 2; } return s.substr(x + 1, count); } string longestPalindrome(string s) { if (s.length() == 0) return ""; if (s.length() == 1) return s; string sub = s.substr(0, 1); for (int k = 0; k < s.length()-1;k++) { string s1 = ishuiwen(s,k,k); if (s1.length()>sub.length()) sub = s1; string s2 = ishuiwen(s, k, k+1); if (s2.length() > sub.length()) sub = s2; } return sub; }};
2、思路二代码(超时):
class Solution94 {//超时public: string longestPalindrome(string s) { if (s.length() == 0) return ""; if (s.length() == 1) return s; int first = 0; int count = 1; for (int k = 0; k < s.length(); k++) { int c_temp = 0; for (int p = s.length() - 1; p > k; p--) { int i = k; if (s[i] != s[p]) continue; int j = p; c_temp = 2; i++; j--; while (i < j) { if (s[i] != s[j]) { c_temp = 0; break; } else { i++; j--; c_temp += 2; } } if (i == j) { c_temp += 1; } if (c_temp > count) { count = c_temp; first = k; } } } return s.substr(first,count); }};
5.4 Python代码:
1、思路一代码(1352ms):
class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ if len(s)==0 or len(s)==1: return s sub=s[0:1] for i in range(len(s)-1): def ishuiwen(s,x,y): while x>=0 and y<len(s) and s[x]==s[y]: x-=1 y+=1 return s[x+1:y] s1=ishuiwen(s,i,i) if len(s1)>len(sub): sub=s1 s2=ishuiwen(s,i,i+1) if len(s2)>len(sub): sub=s2 return sub
- 【leetcode】5. Longest Palindromic Substring(Python & C++)
- LeetCode 5. Longest Palindromic Substring(C++)
- 5.Longest Palindromic Substring Leetcode Python
- [Leetcode]5.Longest Palindromic Substring @ Python
- [leetcode: Python]5.Longest Palindromic Substring
- 【C++】【LeetCode】5. Longest Palindromic Substring
- Leetcode题解(Python): 5. Longest Palindromic Substring
- 【Leetcode】【python】Longest Palindromic Substring
- leetcode Longest Palindromic Substring(C)
- LeetCode 5 Longest Palindromic Substring(C,C++,Python,Java)
- Leetcode 5. Longest Palindromic Substring(字符串)
- (Leetcode)5. Longest Palindromic Substring
- [LeetCode]5 Longest Palindromic Substring(C++,Python实现)
- [LeetCode] 005. Longest Palindromic Substring (Medium) (C++/Java/Python)
- LeetCode 5. Longest Palindromic Substring
- LeetCode --- 5. Longest Palindromic Substring
- LeetCode 5.Longest Palindromic Substring
- [Leetcode] 5. Longest Palindromic Substring
- PHP超级全局变量、魔术变量
- linux root删除文件提示permission denied
- CSS之图片水平垂直居中总结
- Python中int32转int64
- C++定时器的实现与应用
- 【leetcode】5. Longest Palindromic Substring(Python & C++)
- 阿里云服务器apache环境安装ssl证书
- CPU的cache line原理
- 做一个自己的WiFi密码查看器——Android
- 普通Api和RESTful Api的区别
- 职业规划
- 新路程------hi3516a 内置rtc调试
- JAVA基础学习
- QT屏幕保护程序二