leetcode——[069]Sqrt(x)X的平方根

题目

实现 int sqrt(int x) 函数。

计算并返回 x 的平方根，其中 x 是非负整数。

由于返回类型是整数，结果只保留整数的部分，小数部分将被舍去。

示例 1:

输入: 4
输出: 2

示例 2:

输入: 8
输出: 2
说明: 8 的平方根是 2.82842..., 
     由于返回类型是整数，小数部分将被舍去。

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.

解题方法

二分法，取$0$为起点，$x$为终点，每次循环判断中点值$mid$的平方与$x$的大小，$mid^2<= x$，则起点$left+1$；否则终点$right-1$，直到$left>right$。值得注意的是取中点值时用int mid = left + (right - left) / 2防止$int$型数据溢出，如果用int mid = (left + right) / 2，left + right >= 2^32时越界。这段代码跑了6ms，超过了97.63%的Java提交。

public class Solution {
    // Method 1 6ms 97.63%
    public int mySqrt(int x) {
        if (x == 0 || x == 1) {
            return x;
        }
        int ans = 0, left = 0, right = x;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if (mid <= x / mid) {
                left = mid + 1;
                ans = mid;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }
}