Painting solution algorithm: 7 integer reversal

Topic Link

leetcode-cn.com/problems/re…

Title Description

Gives a 32-bit signed integer, you need this integer number on each inverted.

Example 1:

输入: 123
输出: 321
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Example 2:

输入: -123
输出: -321
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Example 3:

输入: 120
输出: 21
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note:

Suppose we have an environment only stores the 32bit signed integer, the range of its value [−231, 231 − 1]. Please According to this hypothesis, if integer overflow after reverse it returns 0.

Problem-solving program

Thinking

• Tags: Mathematics
• If you do not consider this question overflow problem is very simple. There are two ideas to solve the overflow problem, the first idea is to convert a string by Jia try catchto address the way, the second idea is to solve mathematical calculations.
• Due to the lower string conversion efficiency and use more library function, so the program does not consider the problem-solving method, but to solve mathematical calculations.
• By circulating the numbers xof every one apart, in calculating the new value every step of determining whether or not overflow.
• There are two overflow condition, a maximum value is an integer MAX_VALUE, and the other is an integer less than the minimum MIN_VALUE, the current calculation result is provided ans, next to pop.
• From ans * 10 + pop > MAX_VALUEthis point of view overflow condition
  • When there is ans > MAX_VALUE / 10, and 还有pop需要添加when, it must overflow
  • When there is ans == MAX_VALUE / 10, and pop > 7when, it must overflow, 7is 2^31 - 1in single digits
• From ans * 10 + pop < MIN_VALUEthis point of view overflow condition
  • When there is ans < MIN_VALUE / 10, and 还有pop需要添加when, it must overflow
  • When there is ans == MAX_VALUE / 10, and pop < -8when, it must overflow, 8is -2^31in single digits

Code

class Solution {
    public int reverse(int x) {
        int ans = 0;
        while (x != 0) {
            int pop = x % 10;
            if (ans > Integer.MAX_VALUE / 10 || (ans == Integer.MAX_VALUE / 10 && pop > 7)) 
                return 0;
            if (ans < Integer.MIN_VALUE / 10 || (ans == Integer.MIN_VALUE / 10 && pop < -8)) 
                return 0;
            ans = ans * 10 + pop;
            x /= 10;
        }
        return ans;
    }
}
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Painting Solutions

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Reproduced in: https: //juejin.im/post/5d022cc05188255a57151f05