leetcode69 Sqrt(x)

"""
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.
"""
"""
一道easy题被我做的一波三折
"""
"""
第一种解法 遍历找i*i <= x <(i+1)*(i+1)
wrong answer
Input: 2147395599
Output: 99
Expected: 46339
"""
class Solution1:
    def mySqrt(self, x):
        for i in range(100):
            if i*i <= x <(i+1)*(i+1):
                break
        return i
"""
第二种解法开始考虑二分，但
Time Limit Exceeded
"""
class Solution2:
    def mySqrt(self, x):
        i = x // 2
        while i*i != x:   #这里没有抓住本质的停止迭代条件
            if i*i > x:
                i = i // 2
            else:
                i = ((i // 2) + i) // 2
        return i
"""
正确的二分法，left, mid, right
"""
class Solution3:
    def mySqrt(self, x):
        left = 0
        right = x
        while left <= right:
            mid = (left + right) // 2
            s = mid*mid
            if x == s:
                return mid
            elif x > s:
                left = mid + 1
            else:
                right = mid - 1
        return right

"""
高级解法：牛顿迭代法
传送门：https://blog.csdn.net/weixin_42130471/article/details/82730562
r[i] = r[i-1]/2+x/r[i-1]/2
"""
class Solution4:
    def mySqrt(self, x):
        r = x
        while r*r > x:
            r = (r + x//r) // 2 #!!!牛顿迭代方程
        return r