(LeetCode OJ)Reverse Integer【7】

(LeetCode OJ)Reverse Integer【7】

Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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//思路想清楚：要先知道數的正負
//先一位一位的個位取出來(他是新數的最高位，循環一次*10)，取了之後再將這個數縮小10倍（取整）重復如此操作
//題目需要溢出處理
class Solution {
public:
    int reverse(int x) {
        int sgn=-1;
        if(x<0)
			x=x*sgn;
        else
			sgn=1;
		long long ans=0;
		while(x)
		{
			ans*=10;
			ans+=x%10;
			x/=10;
		}
		if(sgn==-1)
		    ans*=-1;
		    
		//溢出判斷
		if(ans>INT_MAX || ans