分类：BinarySearch
考察知识点：BinarySearch 牛顿法
最优解时间复杂度：O(logn)(BinarySearch) O(?)牛顿法
最优解空间复杂度：O(1)
69. 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.
代码：
BinarySearch方法:
class Solution:
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
start=0
end=x
while(end-start>1):
mid=(end-start)//2+start
res=mid*mid
if res==x:
return mid
elif res<x:
start=mid
else:
end=mid
if end*end==x:
return end
else:
return start
牛顿法:
class Solution:
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
res=x
while(res*res>x):
res=(res+x//res)//2
return int(res)
讨论：
1.这道题有两种解法，在面试的时候最好还是把两种解法都写上比较好
2.牛顿法的具体操作参考367题？？？（其实我并不打算刷到那么后面orz）(res+x//res)//2....这是什么神操作啊。。。好厉害
BinarySearch
牛顿法
