# Leetcode 7. Reverse Integer (C語言)
- 題目
Given a 32-bit signed integer, reverse digits of an integer.
- 範例
Example 1:
Input: 123
Output: 321

Example 2:
Input: -123
Output: -321
- Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [$−2^{31}$, $2^{31} − 1$]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

int reverse(int x){
    int y=0;
    while(x!=0){
        if(y>INT_MAX/10||y<INT_MIN/10)return 0;
        else if(y==INT_MAX/10||y==INT_MIN/10){
            if(x%10>7||x%10<-8)return 0;
        }
        y=y*10+x%10;
        x/=10;
    }
    return y;
}

思路:mod(%)可取尾數 搭配除法即可反轉，此題的重點反而在於overflow。
INT_MAX$=2^{31} − 1=2147483647$，INT_MIN$=-2^{31}=-2147483648$。

如果某一次while當中y>214748364(INT_MAX/10)，那麼乘10後必然會overflow。
同理若y<-214748364(INT_MIN/10)也會overflow。

要是y=INT_MAX/10或INT_MIN/10呢? 就要再看尾數(x%10)有沒有>7或<-8來判斷會不會overflow

若INT_MAX/10>y>INT_MAX/10則必然不會overflow。