LeetCode-Easy刷题(15) Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

x is guaranteed to be a non-negative integer.

Example 1:

Input: 4Output: 2

Example 2:

Input: 8Output: 2Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated

实现开平方这个方法,返回整数部分.

//这题应该用二分法来查找    //二分法    public static int mySqrt(int x) {        if(x==0 || x ==1){            return x;        }        int left  = 0;        int right = x;        while(left<=right){            int mid = (left + right)/2;            if(x/mid==mid && x%mid==0){                return mid;            }            if(x/mid<mid){                right = mid -1;            }else{                left = mid + 1;            }        }        return right;    }