LeetCode 69 _ Sqrt(x) 求平方根 (Easy)

Description:

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.

Solution:

Code:

public int mySqrt(int x) {
    int left = 0, right = x;
    if (x <= 1){
    	return x;
    }
    while (left < right){
    	int mid = left + (right - left)/2;
    	// mid写左半以免越界!
    	if (x / mid >= mid){ // 如果x大于mid^2,证明mid太小,要取右半
    		left = mid + 1; // 如果x刚好等于^2,也将其+1,便于之后取-1的结果
    	}else{
    		right = mid;     // 如果x小于mid^2,证明mid太大,要取左半
    	}
    }
    return right-1;
}

提交情况:

Runtime: 1 ms, faster than 100.00% of Java online submissions for Sqrt(x).
Memory Usage: 32.3 MB, less than 100.00% of Java online submissions for Sqrt(x).

原文地址：https://www.cnblogs.com/zingg7/p/10679062.html