Reverse Integer
Category Difficulty Likes Dislikes algorithms Medium (26.18%) 5870 8734 Tags Companies Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.
Assume the environment does not allow you to store 64-bit integers (signed or unsigned).
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Example 1:
Input: x = 123
Output: 321
Example 2:
Input: x = -123
Output: -321
Example 3:
Input: x = 120
Output: 21
Example 4:
Input: x = 0
Output: 0
Constraints:
-231 <= x <= 231 - 1
C++ Solution
思路是首先不断除10获取低位的数字，然后通过FIFO的数据结构，从头部拿出并累计成新数字。最大的问题是当新数字超过INT的范围时，需要返回0，这里的做法是用更大的类型去进行计算，然后通过INT_MAX和INT_MIN比较判断。
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class Solution {
public:
int reverse(int x) {
list <int> myList;
while (x) {
int digit = x % 10;
x = x / 10;
myList.push_back(digit);
}
long ans = 0;
while (!myList.empty()) {
ans = ans * 10 + myList.front();
myList.pop_front();
}
if (ans > INT_MAX || ans < INT_MIN) {
return 0;
}
return ans;
}
};
