69. Sqrt(x)

Posted 2022-12-01 wentiliangkaihua

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

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;

 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.