給出一個 32 位的有符號整數，你需要將這個整數中每位上的數字進行反轉。

• 示例 1:

輸入: 123
輸出: 321

• 示例 2:

輸入: -123
輸出: -321

• 示例 3:

輸入: 120
輸出: 21

• 注意:

假設我們的環境只能存儲得下 32 位的有符號整數，則其數值范圍為 [?231, 231 ? 1]。請根據這個假設，如果反轉后整數溢出那么就返回 0。

class Solution {
public:int reverse(int x) {int max = (1<<31)-1;int min = (1<<31);int numbers[10];bool label = true;if (x < 0) {label = false;if (x != min) {x = -x;} else {return 0;}} if (x == 0) {return 0;}int count = 0;while (x > 0) {int tmp = x % 10;x /= 10;numbers[count++] = tmp;}long result = 0;for (int i = 0; i < count; i++) {result = result * 10 + numbers[i];}if (result > max) {return 0;}if (!label) {return -result;} else {return result;}}
};

or

class Solution {
public:int reverse(int x) {bool negative = false;if (x < 0) {negative = true;if (x == INT_MIN) {return 0;} else {x = -x;}}long result = 0;while (x > 0) {result = result * 10 + x%10;x/=10;}if (result > INT_MAX) {return 0;} else if (negative){return -result;} else {return result;}}
};

or

class Solution {
public:int reverse(int x) {long result = 0;while (x != 0) {result = result * 10 + x%10;x/=10;}if (result > INT_MAX || result < INT_MIN) {return 0;} else {return result;}}
};

注：

INT_MIN 和 INT_MAX，均定義在 <limits.h>頭文件中#define INT_MAX 2147483647
#define INT_MIN (-INT_MAX - 1)

來源：力扣（LeetCode）