69. Sqrt(x)(二分查找)

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.

 1 class Solution {
 2 public:
 3     int mySqrt(int x) {
 4         if (x <=1) return x;
 5         int low = 0;
 6         int high = x ;
 7         while(low <= high) {
 8             int mid = low + (high-low)/2;
 9             if (x/mid > mid) {
10                 low = mid + 1;
11             } else if(x/mid < mid) {
12                 high = mid - 1;
13             } else {
14                 return mid;
15             }
16         }
17         // 正常二分法，如果没找到，返回的是大于target 的位置。此题需要返回小于target的位置。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。
18         return low - 1;
19     }
20 };