「每日一道算法题」Reverse Integer

Algorithm

OJ address

Leetcode website : 7. Reverse Integer

Description

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2的31次方, 2的31次方 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution in C (one)

int reverse(int x) {
    int t = 0;
    if (x == 0) return 0;
    while (!x%10) {
        x/=10;
    }
    int sum = 0;
    long long reallysum = 0;
    while (x) {
        reallysum *=10;
        sum *= 10;
        t = x%10;
        reallysum+=t;
        sum+=t;
        x/=10;
    }
    if (reallysum != sum) return 0;
    return sum;
}

Solution in C (Two)

int reverse(int x) {
    int t = 0;
    if (x == 0) return 0;
    while (!x%10) {
        x/=10;
    }
    int sum = 0;
    int tmp = 0;
    while (x) {
        sum *= 10;
        if (sum/10 != tmp) return 0;
        t = x%10;
        sum+=t;
        x/=10;
        tmp = sum;
    }
    return sum;
}

My Idea

题目含义是，给定一个int类型的整数，然后进行数字反转，输出。

1. 反转后，将前导0去掉，例如2300 -> 0023 ->23
2. 如果超过 INT_MAX , 或者小鱼 INT_MIN，则输出0，关于这个如何判断，有两种简单的方法，第一种方法是用long long来存取变量，如果大于INT_MAX或者小于INT_MIN，则输出0.第二种方法就是如果超出最大值，或小于最小值，则你最高位后面的尾数是会因为超出最大值而跟着改变的，所以你只要检测尾数如果变化，就输出0即可，这就是我代码里的第二种方法。