zoukankan      html  css  js  c++  java
  • [LeetCode]Longest Palindromic Substring

    题目:Longest Palindromic Substring

    查找最长回文子串

    思路:

    一个指针从头部,一个指针从尾部,对比每一个字母,若相等则可能是回文子串,则,检测子串是否回文,是则比较和已知的子串长度,更长就记录其起始和终止坐标,否则就放弃。

     1 /**********************************************************************
     2 Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
     3 Example:
     4 Input: "babad"
     5 Output: "bab"
     6 Note: "aba" is also a valid answer.
     7 Example:
     8 Input: "cbbd"
     9 Output: "bb"
    10 **********************************************************************/
    11 #include <stdio.h>
    12 #include <ctype.h>
    13 
    14 char* longestPalindrome(char* s) {
    15     int length = strlen(s);
    16     int i,j,max = 1,maxi = 0,maxj = 0,flag = 1;//flag标志是否是回文子串
    17     for(i = 0;i < length && i + max < length;i++){//从头开始检测,最后剩余未检测长度少于一直最长解则放弃
    18         for(j = length - 1;j > i;j--){//从尾部开始检测
    19             flag = 1;
    20             if(s[i] == s[j]){//可能回文
    21                 int head = i + 1,tail = j - 1;
    22                 while(head < tail){//检测是否是回文子串
    23                     if(s[head] != s[tail]){
    24                         flag = 0;
    25                         break;
    26                     }
    27                     head++;
    28                     tail--;
    29                 }
    30                 if(flag == 1){//如果是回文子串
    31                     if(max < j - i + 1){//长度是否比已知最优解长
    32                         max = j - i + 1;
    33                         maxi = i;
    34                         maxj = j;
    35                     }
    36                     break;//继续收缩尾部指针,也得不到更优的解,所以放弃
    37                 }
    38             }
    39         }
    40     }
    41 
    42     char *substr = (char *)malloc((max + 1)*sizeof(char));
    43     memset(substr,0,(max + 1)*sizeof(char));
    44     strncpy(substr,s + maxi,max);
    45     return substr;
    46 }
    47 
    48 void main(){
    49     char s[] = "babadada";
    50     char *sp = longestPalindrome(s);
    51     printf("%s
    ",sp);
    52     free(sp);
    53 }

    上面的思路是从两边向中间收束,另一个思路是从中间向两边发散。

    具体如下:

    先找当前下标为中心的回文子串,比较它和最优解,更优则更新最优解,

    递归查找中心点向左和向右偏移后的回文子串。

     1 /**********************************************************************
     2 Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
     3 Example:
     4 Input: "babad"
     5 Output: "bab"
     6 Note: "aba" is also a valid answer.
     7 Example:
     8 Input: "cbbd"
     9 Output: "bb"
    10 **********************************************************************/
    11 #include <stdio.h>
    12 #include <ctype.h>
    13 
    14 int max = 1;//记录已知最长的回文子串的长度
    15 int maxi = 0;//记录已知最长的回文子串的头部坐标
    16 int maxj = 0;//记录已知最长的回文子串的尾部坐标
    17 
    18 void longestPSubstring(char *s,int i){//记录当前搜索的位置
    19     int head = i - 1;
    20     int tail = i + 1;
    21     int length = strlen(s);
    22     while((head >= 0 || tail < length) && s[head] == s[tail]){//检测以i为中心的回文子串
    23         head--;
    24         tail++;
    25     }
    26     if(tail - head - 1 > max){//比较以i为中心的回文子串是否更长
    27         maxi = head + 1;
    28         maxj = tail - 1;
    29         max = tail - head - 1;
    30     }
    31     
    32     if(i - 1 > 0){
    33         longestPSubstring(s,i - 1);//向左偏移
    34     }
    35     if(i + 1 < length - 1){
    36         longestPSubstring(s,i + 1);//向右偏移
    37     }
    38 }
    39 
    40 char* longestPalindrome(char* s) {
    41         int length = strlen(s);
    42     longestPSubstring(s,length/2);
    43 
    44     char *substr = (char *)malloc((max + 1)*sizeof(char));
    45     memset(substr,0,(max + 1)*sizeof(char));
    46     strncpy(substr,s + maxi,max);
    47     return substr;
    48 }
    49 
    50 void main(){
    51     char s[] = "babadada";
    52     char *sp = longestPalindrome(s);
    53     printf("%s
    ",sp);
    54     free(sp);
    55 }
  • 相关阅读:
    9.1做JS的题目(2)
    9.1做JS的题目
    8.31做JS的题目
    8.30做JS的题目
    扫码跳转微信小程序(服务端微信API生成二维码)
    扫码跳转微信小程序(微信公众平台配置测试二维码)
    项目配置:maven下载与配置、tomcat下载与配置
    Java基础:常用工具_API
    Java基础: 抽象类、接口、final关键字、static关键字
    java基础: 封装、继承、多态
  • 原文地址:https://www.cnblogs.com/yeqluofwupheng/p/6666867.html
Copyright © 2011-2022 走看看