LeetCode 69 x square root of Newton iteration of binary search

Title Description

Implement int sqrt(int x)functions.

Computes and returns x square root, wherein x is a non-negative integer.

Since the return type is an integer, the integer part of the result to retain only the fractional part is rounded down.

Example 1:

输入: 4
输出: 2

Example 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。

analysis

The square root of self-realization. Methods as below:

Cycle of violence

class Solution {
public:
    int mySqrt(int x) {
			long long res = 0;
			for (long long i = 0; i < x; i++) {
					if ((i * i) == x) {res = i; break;}
					long long tmp = i + 1;
					if ((tmp * tmp) > x) {res = i; break;}
					if ((tmp * tmp) == x) {res = tmp; break;} 
			}
			return res;
    }    

Library Functions

C++: return sqrt(x);

Java: return int(math.sqrt(x))

Binary search

class Solution {
public:
    int mySqrt(int x) {
      	 long long begin = 0;
         long long end = x / 2 + 1;
         long long mid;
         long long res;
         while (begin <= end) {
             mid = (begin + end) / 2;
             res = mid * mid;
             if (res == x) return mid;
             if (res < x) begin = mid + 1;
             if (res > x) end = mid -1;
         }
        return end;
}  

Newton iteration

class Solution {
public:
    int mySqrt(int x) {
				if (x == 0) return 0;
        double flag = 0;
        double res = 1;
        while (res! = flag)
        {
            flag = res;
            res = (res + x / res ) / 2;
        }
        return int(res);
    }
};

reference

Newton iterative learning:
https://www.cnblogs.com/grandyang/p/4346413.html
https://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html