LeetCode - Sqrt(x)

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.

思路：用二分法来做，average时间复杂度 O(logn)

class Solution {
    public int mySqrt(int x) {
    
        if (x<2) return x;
        
        int l = 0;
        int h = x;
        
        int res = 0;
        while (l <= h) {
            int m = l + (h-l)/2;
            if (m <= x/m && (m+1) > x/(m+1)){
                return m;
            } 
            if (m > x/m) {
                h = m-1;   
            }
            else if (m < x/m) {
                res = m;
                l = m+1;
            }
        }
        return res;
    }
}

 

class Solution {
    public int mySqrt(int x) {
        if (x == 0) return 0;
        if (x <= 3) return 1;
        
        int left = 2;
        int right = x/2;
        int res = 0;
        while(left <= right) {
            int mid = left + (right - left)/2;
            if (mid <= x/mid && (mid + 1) > x/(mid + 1)) {
                return mid;
            }
            if (mid > x/mid) {
                right = mid - 1;
            }
            else {
                left = left + 1;
            }
        }
        return 0;
    }
}