题目地址：https://leetcode.com/problems/sqrtx/description/
题目描述
Implement int sqrt(int x).
Compute and return the square root of x.
xis guaranteed to be a non-negative integer.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated.
题目大意
求x的算术平方根向下取整。
解题方法
方法一：库函数
求算数平方根。可以直接用库。
import math
class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
return int(math.sqrt(x))
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方法二：牛顿法
牛顿法详见：https://en.wikipedia.org/wiki/Newton%27s_method
这个问题其实就是求f(x)=num - x ^ 2的零点。
那么，Xn+1 = Xn - f(Xn)/f'(Xn).
又f'(x) = -2x.
得Xn+1 = Xn +(num - Xn ^ 2)/2Xn = (num + Xn ^ 2) / 2Xn = (num / Xn + Xn) / 2.
即t= (num / t + t) / 2.
class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
num = x
while num * num > x:
num = (num + x / num) / 2
return num
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方法三：二分查找
这个题是二分查找的经典题目了，直接套用二分查找的模板即可。这里贡献一个二分查找的模板，模板中查找的区间是[l, r)，即左闭右开。
def binary_searche(l, r):
while l < r:
m = l + (r - l) // 2
if f(m): 判断找了没有，optional
return m
if g(m):
r = m new range [l, m)
else:
l = m + 1 new range [m+1, r)
return l or not found
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这个题的二分查找版本的代码如下：
class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
left, right = 0, x + 1
[left, right)
while left < right:
mid = left + (right - left) // 2
if mid ** 2 == x:
return mid
if mid ** 2 < x:
left = mid + 1
else:
right = mid
return left - 1
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二刷的时候用了二分查找，但是写法和上面略有区别。
class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
left, right = 0, x + 1[left, right)
while left < right:
mid = (left + right) // 2
if mid ** 2 == x:
return mid
elif (mid - 1) ** 2 < x and mid ** 2 >= x:
return mid - 1
elif mid ** 2 < x:
left = mid + 1
else:
right = mid
return left
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如果是C++就比较痛苦了，因为要考虑到数字是不是越界了，所以我用了long long.
class Solution {
public:
int mySqrt(long long x) {
if (x == 0) return 0;
long long left = 0, right = x + 1; //[left, right)
while (left < right) {
long long mid = left + (right - left) / 2;
if (mid == x / mid) {
return mid;
} else if (mid < x / mid) {
left = mid + 1;
} else {
right = mid;
}
}
return left - 1;
}
};
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如果只使用int的话，需要考虑right = x + 1这一步可能会超边界，所以仍然使用[left,right)区间，那么当x<=1的时候，返回的应该是x。
class Solution {
public:
int mySqrt(int x) {
if (x <= 1) return x;
int left = 0, right = x; //[left, right)
while (left < right) {
int mid = left + (right - left) / 2;
if (mid == x / mid) {
return mid;
} else if (mid < x / mid) {
left = mid + 1;
} else {
right = mid;
}
}
return left - 1;
}
};
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