[日常刷题]leetcode第三天

9. Palindrome Number

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Coud you solve it without converting the integer to a string?

Solution in C++:

方法一：将数字从尾倒头颠倒后进行比较

bool isPalindrome(int x) {
       // 负数
       if ( x < 0)
           return false;
       else{
           int y = 0;
           int tmp = x;
           while(x)
           {
               y = y * 10 + x % 10;
               x = x / 10;
           }
           cout << "y = " << y << endl;
           if (y == tmp)
               return true;
           else
               return false;
       }
   }

方法二：将数字一半进行倒转后进行比较
关键点：1.剔除不符合条件（负数、除0外末尾为0的数）
2.如何判断是否倒转了一半（当reverseNumber > originalNumber）

 bool isPalindrome(int x) {
     // 负数
     if ( x < 0 || (x % 10 == 0 && x != 0))
         return false;
     else{
         int y = 0;
         while(x > y)
         {
             y = y * 10 + x % 10;
             x = x / 10;
         }
         if (y == x || y / 10 == x)
             return true;
         else
             return false;
     }
 }

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13. Roman to Integer

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9.
• X can be placed before L (50) and C (100) to make 40 and 90.
• C can be placed before D (500) and M (1000) to make 400 and 900.
  Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Solution in C++：

int romanToInt(string s) {
        map<char, int> table = {{'I', 1},{'V',5},
                                {'X',10},{'L',50},
                                {'C',100},{'D',500},{'M',1000}};
        size_t size = s.size();
        int result = 0;
        for ( int i = 0; i < size; ++i)
        {
            if (i == size - 1)
            {
                result += table[s[i]];
            }
            else if (table[s[i]] < table[s[i+1]])  // 特殊情况
            {
                result += table[s[i+1]] - table[s[i]];
                ++i;
            } else{
                result += table[s[i]];
            }
        }
        return result;
    }

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小结：

今天做的题目都比较简单，但是写出来的程序并不能不调试就直接提交。

1. 逻辑控制部分还是存在不足，比如第13题中for循环的控制处的条件边界以及分情况+1 +2等思考略有不足。
2. 第9题中，折半有想到但是并没有坚持做下去，还是采用了暴力的方法。