题目链接 : https://leetcode-cn.com/problems/sqrtx/
题目描述:
实现 int sqrt(int x)
函数。
计算并返回 x 的平方根,其中 x 是非负整数。
由于返回类型是整数,结果只保留整数的部分,小数部分将被舍去。
示例:
示例 1:
输入: 4
输出: 2
示例 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。
思路:
思路一:库函数
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
return int(math.sqrt(x))
思路二: 二分法
思路三: 牛顿法.
代码:
思路二:
class Solution:
def mySqrt(self, x: int) -> int:
left = 0
right = math.ceil(x / 2)
res = 0
while left <= right:
mid = left + (right - left) // 2
tmp = mid * mid
if tmp == x:
return mid
elif tmp < x:
left = mid + 1
else:
right = mid - 1
return right
java
class Solution {
public int mySqrt(int x) {
int left = 1;
int right = (x / 2) + 1;
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 - 1;
}
return right;
}
}
思路三:
class Solution:
def mySqrt(self, x: int) -> int:
r = x
while r * r > x:
r = (r + x // r) // 2
return r
java
class Solution {
public int mySqrt(int x) {
long r = x;
while (r * r > x) r = (r + x / r) / 2;
return (int) r;
}
}