原题传送门

1. 题目描述

2. Solution

1、思路分析
二分查找，在[0, x]这个区间内执行二分查找，求mid，使得mid * mid = x。

2、代码实现

package Q0099.Q0069sqrtx;


/*
  A Binary Search Solution from wikipedia:
  https://en.wikipedia.org/wiki/Integer_square_root
 */
public class Solution {

    public int mySqrt(int x) {
        if (x == 0) return 0;
        long lo = 0, hi = x;
        while (lo < hi) {
            long mid = (lo + hi + 1) / 2;
            if (mid * mid > x) {
                hi = mid - 1;
            } else {
                lo = mid;
            }
        }
        return (int) lo;
    }
}

3、复杂度分析
时间复杂度: O(log n)
空间复杂度: O(1)

3. Soution

1、思路分析
牛顿法

2、代码实现

package Q0099.Q0069sqrtx;

/*
  Algorithm using Newton's method, from wikipedia
  https://en.wikipedia.org/wiki/Integer_square_root
 */
public class Solution1 {

    public int mySqrt(int x) {
        long n = x;
        while (n * n > x)
            n = (n + x / n) / 2;
        return (int) n;
    }
}

3、复杂度分析
时间复杂度: O(log x)
空间复杂度: O(1)

标签：int,lo,复杂度,wikipedia,mid,Sqrt,hi,LeetCode,0069
来源： https://www.cnblogs.com/junstat/p/16184287.html