LeetCode 69.x square root (simple) divide and conquer and Newton iteration

topic

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, only the integer part of the result is retained, and the decimal part will be discarded.

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。

• Binary search

class Solution{
    
    
    public int mySqrt(int x){
    
    
        if(x == 0 || x == 1){
    
    
            return x;
        }
        int left = 1;
        int right = x;
        while(left < right){
    
    
            int mid = left + (right-left)/2;
            if (mid * mid > x){
    
    
                right = mid - 1;
            }else{
    
    
                left = mid + 1;
            }
        }
        return right;                  
    }
}

mid = left + (right-left)/2, Writing this way is afraid that left+rightif the value is too large in a strongly typed language , it may cross the boundary, so use this technique to deal with it.

• Newton iteration method

class Solution：
	def mySqrt(self,x):
        if x < 0 :
            raise Exception('不能输入负数')
         if x == 0:
            return 0
        #起始的时候在1，随意设置
        cur = 1
        while True:
            pre = cur
            cur = (cur + x / cur) / 2
            if abs(cur-pre) < 1e-6:
                return int(cur)

• Newton iteration to simplify the code

class Soluction(object):
	def mySqrt(self,x):
        r=x
        while r*r>x:
            r=(r+x/r)/2
		return r