367. Valid Perfect Square

https://leetcode.com/problems/valid-perfect-square/description/
Solution 1. For 32 bit integer input num, sqrt(num) < pow(2,16). Let n=0, Start from 15th bit, n = n + pow(2,15). If num < n*n, decreasing n to pow(2,14), pow(2,13), ... until num > n*n, say n = pow(2,i). Then start again from i-1 bit.
     bool isPerfectSquare(int num) {
        vector<int> sq;
        int n = 0;
        for(int i=15; i>=0; i--){
            while(pow(n+(1<<i),2) > num && i>=0) i--;
            n += (1<<i);
            if(pow(n,2) == num) return true;
        }
        return false;
    }
Solution 2. 1 + 3 + 5 + ... + (2n-1) = n*n
     bool isPerfectSquare(int num) {
        int i = 1;
        while (num > 0) {
            num -= i;
            i += 2;
        }
        return num == 0;
    }