zoukankan      html  css  js  c++  java
  • LeetCode 125. Valid Palindrome

    https://leetcode.com/problems/valid-palindrome/

    Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

    For example,
    "A man, a plan, a canal: Panama" is a palindrome.
    "race a car" is not a palindrome.

    Note:
    Have you consider that the string might be empty? This is a good question to ask during an interview.

    For the purpose of this problem, we define empty string as valid palindrome.

    • 字符串题。两种解法,一种是直接把string翻转比较,另一种是传统的首尾两个指针移动比较。
    • 注意对指针-- end,end不能为size_t类型, 否则一旦被减为-1则会被转变为INT_MAX,程序进入死循环。
    • 学习了string相关STL算法和函数,往string里添加char用push_back。
    • toupper - C++ Reference
      •   http://www.cplusplus.com/reference/cctype/toupper/
    • isalnum - C++ Reference
      •   http://www.cplusplus.com/reference/cctype/isalnum/?kw=isalnum
    • transform - C++ Reference
      •   http://www.cplusplus.com/reference/algorithm/transform/?kw=transform
    • string::push_back - C++ Reference
      •   http://www.cplusplus.com/reference/string/string/push_back/
    • reverse_copy - C++ Reference
      •   http://www.cplusplus.com/reference/algorithm/reverse_copy/
    • reverse_copy
      •   https://msdn.microsoft.com/zh-cn/library/c7ty5c36.aspx
     1 #include <iostream>
     2 #include <string>
     3 #include <cctype>  // toupper, isalnum
     4 #include <algorithm>  // transform
     5 using namespace std;
     6 
     7 class Solution {
     8 public:
     9     bool isPalindrome(string s) 
    10     {
    11          string sTemp = "", sReverse = "";
    12 
    13          // Convert characters to UPPER
    14          transform(s.begin(), s.end(), s.begin(), ::toupper);
    15 
    16          // considering only alphanumeric characters
    17          for (int i = 0; i < s.length(); i ++)
    18              if (isalnum(s.at(i)))
    19                 sTemp.push_back(s.at(i)); // Appends character to the end of the string, increasing its length by one.
    20 
    21          // Need to allocate space if use reverse_copy
    22          sReverse.resize(sTemp.length());
    23          reverse_copy(sTemp.begin(), sTemp.end(), sReverse.begin());
    24 /*
    25          sReverse = sTemp;
    26          reverse(sReverse.begin(), sReverse.end());
    27 */         
    28 /*
    29          cout << sTemp << endl;
    30          cout << sReverse << endl;
    31 */         
    32          return sTemp == sReverse;
    33     }
    34     
    35     bool isPalindrome2(string s) 
    36     {
    37          int start = 0, end = s.length() - 1;
    38 
    39          while (start < end)
    40          {
    41                if ( !isalnum(s.at(start)) ) 
    42                   start ++;
    43                else if ( !isalnum(s.at(end)) ) 
    44                     end --;
    45                else if (toupper(s.at(start)) != toupper(s.at(end))) // Remember toupper 
    46                       return false;
    47                else
    48                {
    49                    // Remember to move ptr
    50                    start ++;
    51                    end --;
    52                }
    53          }
    54          
    55          return true;
    56     }
    57     
    58 };
    59 
    60 int main ()
    61 {
    62     Solution testSolution;
    63     string sTest[] = {"A man, a plan, a canal: Panama", "race a car"};
    64     
    65     for (int i = 0; i < 2; i ++)
    66         cout << testSolution.isPalindrome(sTest[i]) << endl;
    67     
    68     getchar();
    69     
    70     return 0;
    71 }
    View Code
  • 相关阅读:
    屏幕截图 从安卓模拟器中识别出屏幕文字
    srcset
    Bitwise and Bit Shift Operators 位运算 取反 补码
    text recognizer (OCR) Engine 光学字符识别
    删除目录下 某类名字的文件
    appmaptile
    登录框
    将代码设置的剪切板内容通过输入法软件粘贴入app搜索框
    面向问题的高级语言
    使用心理视觉来进行图像处理
  • 原文地址:https://www.cnblogs.com/pegasus923/p/5612541.html
Copyright © 2011-2022 走看看