# 3 JAVA solutions with explanation

• The three solutions are as the follows, solution1 and solution3 are pretty straight forward.

`````` Look for the critical point: i * i <= x && (i+1)(i+1) > x
``````

A little trick is using i <= x / i for comparison, instead of i * i <= x, to avoid exceeding integer upper limit.

Solution1 - Binary Search Solution: Time complexity = O(lg(x)) = O(32)=O(1)

``````public int mySqrt(int x) {
if (x == 0) return 0;
int start = 1, end = x;
while (start < end) {
int mid = start + (end - start) / 2;
if (mid <= x / mid && (mid + 1) > x / (mid + 1))// Found the result
return mid;
else if (mid > x / mid)// Keep checking the left part
end = mid;
else
start = mid + 1;// Keep checking the right part
}
return start;
}
``````

Solution2 - Newton Solution: Time complexity = O(lg(x))

I think Newton solution will not be faster than Solution1(Binary Search), because i = (i + x / i) / 2, the two factors i and x / i are with opposite trends. So time complexity in the best case is O(lgx).

Anyone can give the accurate time complexity? Appreciate it!

``````public int mySqrt(int x) {
if (x == 0) return 0;
long i = x;
while(i > x / i)
i = (i + x / i) / 2;
return (int)i;
}
``````

Solution3 - Brute Force: Time complexity = O(sqrt(x))

``````public int mySqrt(int x) {
if (x == 0) return 0;
for (int i = 1; i <= x / i; i++)
if (i <= x / i && (i + 1) > x / (i + 1))// Look for the critical point: i*i <= x && (i+1)(i+1) > x
return i;
return -1;
}``````

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