141. Sqrt(x) 【easy】

141. Sqrt(x) 【easy】

Implement int sqrt(int x).

Compute and return the square root of x.

Example

sqrt(3) = 1

sqrt(4) = 2

sqrt(5) = 2

sqrt(10) = 3

Challenge 

O(log(x))

解法一：

 class Solution {
 public:
     /*
      * @param x: An integer
      * @return: The sqrt of x
      */
     int sqrt(int x) {
         // write your code here
         long start = ;
         long end = x;

         while (start +  < end) {
             long mid = start + (end - start) / ;

             if (mid * mid == x) {
                 return mid;
             }
             else if (mid * mid < x) {
                 start = mid;
             }
             else if (mid * mid > x) {
                 end = mid;
             }
         }

         if (end * end <= x) {
             return end;
         }
         else {
             return start;
         }
     }
 };

注意：

1、start、end、mid用long类型防止溢出

2、最后判断end * end与x的关系时要考虑等号，否则输入数值为0的时候就错了。

解法二：

 class Solution {
 public:
     /**
      * @param x: An integer
      * @return: The sqrt of x
      */
     int sqrt(int x) {
         // write your code here
         long left = ;
         if (x == )
             return ;
         long right = x;
         long mid = left + (right - left) / ;
         while (left +  < right) {
             if (x > mid * mid) {
                 left = mid;
             } else if (x < mid * mid) {
                 right = mid;
             } else {
                 return mid;
             }
             mid = left + (right - left) / ;
         }
         return left;
     }
 };

这里还是两头都判断一下比较保险。