c++-c++编程基础之leetcode题解第7题整数反转.zip

# [7. Reverse Integer (Easy)](https://leetcode.com/problems/reverse-integer/) <p>Given a 32-bit signed integer, reverse digits of an integer.</p> <p><strong>Example 1:</strong></p> <pre><strong>Input:</strong> 123 <strong>Output:</strong> 321 </pre> <p><strong>Example 2:</strong></p> <pre><strong>Input:</strong> -123 <strong>Output:</strong> -321 </pre> <p><strong>Example 3:</strong></p> <pre><strong>Input:</strong> 120 <strong>Output:</strong> 21 </pre> <p><strong>Note:</strong><br> Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹,&nbsp; 2^31&nbsp;− 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.</p> **Related Topics**: [Math](https://leetcode.com/tag/math/) **Similar Questions**: * [String to Integer (atoi) (Medium)](https://leetcode.com/problems/string-to-integer-atoi/) * [Reverse Bits (Easy)](https://leetcode.com/problems/reverse-bits/) ## Solution 1. ```cpp // OJ: https://leetcode.com/problems/reverse-integer/ // Author: github.com/lzl124631x // Time: O(1) // Space: O(1) class Solution { public: int reverse(int x) { if (x == INT_MIN) return 0; int r = 0, sign = x >= 0 ? 1 : -1, y = sign * x, p = 1; while (y) { y /= 10; if (y) p *= 10; } x = sign * x; while (x) { int d = x % 10; x /= 10; if ((INT_MAX - r) / p < d) return 0; r += p * d; p /= 10; } return sign * r; } }; ``` ## Solution 2. ```cpp // OJ: https://leetcode.com/problems/reverse-integer/ // Author: github.com/lzl124631x // Time: O(1) // Space: O(1) class Solution { public: int reverse(int x) { int r = 0; while (x) { int d = x % 10; x /= 10; if ((r >= 0 && (r > INT_MAX / 10 || INT_MAX - 10 * r < d)) || (r < 0 && (r < INT_MIN / 10 || INT_MIN - 10 * r > d))) return 0; r = r * 10 + d; } return r; } }; ```