Leetcode7. Reverse Integer
题目:
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 hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
题目分析:这题主要的难点在于,32bit,超过32bit就返回0。所以这题我有两种方法,第一种,把结果变成longlong类型,判断结果是否在INT32_MAX和INT32_MIN之内。第二种,判读结果啥时候会溢出。
代码:
class Solution {
public:
int reverse(int x) {
int sum = 0;
int n;
while (x != 0) {
if (sum > INT_MAX/10 || sum < INT_MIN/10) {
return 0;
}
n = x%10;
sum = sum * 10 + n;
x /= 10;
}
return sum;
}
};
本文针对LeetCode 7题“整数反转”进行解析,介绍了一种通过循环实现数字反转的方法,并考虑了32位带符号整数范围内的溢出问题。文章提供了C++代码示例,展示了如何安全地处理整数反转操作。
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