LeetCode Algorithm 0007 - Reverse Integer (Easy)

版权声明：本文为博主原创文章，转载请注明出处。 https://blog.csdn.net/darkrabbit/article/details/82791230

LeetCode Algorithm 0007 - Reverse Integer (Easy)

Problem Link: https://leetcode.com/problems/reverse-integer/description/

---

---

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: $[-2^{31}, 2^{31} - 1]$ . For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

---

Solution C++

#pragma once

#include "pch.h"

// Problem: https://leetcode.com/problems/reverse-integer/description/

namespace P7ReverseInteger
{
    class Solution
    {
        public:
        int reverse(int x)
        {
#if true // 使用 string
            string s = x < 0 ? to_string(-x) : to_string(x);
            std::reverse(s.begin(), s.end());
            long long result = x < 0 ? -stoll(s) : stoll(s);
            if (result > numeric_limits<int>::max() || result < numeric_limits<int>::min())
            {
                return 0;
            }
            return (int)result;
#endif

#if false // 使用 number: 网站答案的源码： Pop and Push Digits & Check before Overflow
            int rev = 0;
            while (x != 0)
            {
                int pop = x % 10;
                x /= 10;
                if (rev > INT_MAX / 10 || (rev == INT_MAX / 10 && pop > 7)) return 0;
                if (rev < INT_MIN / 10 || (rev == INT_MIN / 10 && pop < -8)) return 0;
                rev = rev * 10 + pop;
            }
            return rev;
#endif
        }
    };
}

---