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LeetCode

Approach 1: using long

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class Solution {
    public int reverse(int x) {
        long ans = 0;
		while(x != 0){
			ans= ans*10;
			ans+= x% 10;
			x= x/10;
		}

		if (ans > Integer.MAX_VALUE || ans< Integer.MIN_VALUE) return 0; 		
		else return (int)ans;
    }
}

Approach 2: not using long (explanation)

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class Solution {
    public int reverse(int x) {
        int res=0;
        while(x!=0){
            int tail=x%10;
            int newRes=res*10+tail;
            if((newRes-tail)/10!=res) return 0;
            res=newRes;
            x/=10;
        }
        return res;
    }
}

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