LeetCode——Reverse Integer

LeetCode——Reverse Integer

# 7

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.

整数的反转，思路是很简单的，难点是考虑反转之后的溢出。可以使用long long型数据。

C++

class Solution {
public:
int reverse(int x) {
long long res = 0;
bool isPositive = true;
if(x < 0) {
isPositive = false;
x = x * -1;
}
while(x > 0) {
res = res * 10 + x % 10;
x = x / 10;
}
if(res > INT_MAX)
return 0;
if(isPositive)
return res;
else
return -res;
}
};

另一种简洁的做法，是不考虑符号位。

-C++

class Solution {
public:
int reverse(int x) {
int res = 0;
while (x != 0) {
if (abs(res) > INT_MAX / 10) return 0;
res = res * 10 + x % 10;
x /= 10;
}
return res;
}
};

也可以将两种结合在一起。

-C++

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