[leetcode]69. Sqrt(x)开方

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.

题意：

整数开方

Solution1： Binary Search

code

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