7-整数反转

LeetCode 7. Reverse Integer

LeetCode-7

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 [-2^31, 2^31 - 1], then return 0.

method

INT_MIN = -21,4748,3648
INT_MAX =  21,4748,3647

最大最小数都是十位数，如果一个9位数已经比INT_MAX / 10 = 21,4748,364还大了，那再加任何一位数都会溢出
虽然等于INT_MAX / 10也可能会溢出，比如21,4748,3649，但是注意原数是在整数范围内的，所以并不会出现这种情况，最大也只能是1463847412

• 要先判断再加数

int reverse(int x) {
    int num = 0;
    while (x) {
        if (num < (INT_MIN / 10) || num > (INT_MAX / 10)) return 0;
        num = num * 10 + x % 10;
        x /= 10;
    }
    return num;
}