daily leetcode - sqrtx - !
题目地址
https://leetcode.com/problems/sqrtx/
题目描述
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.
思路
这道题要求我们求平方根,我们能想到的方法就是算一个候选值的平方,然后和x比较大小,为了缩短查找时间,我们采用二分搜索法来找平方根,这里属于博主之前总结的LeetCode Binary Search Summary 二分搜索法小结中的第三类的变形,找最后一个不大于目标值的数.
关键点解析
代码
解法一:
class Solution {
public:
int mySqrt(int x) {
if (x <= 1) return x;
int left = 0, right = x;
while (left < right) {
int mid = left + (right - left) / 2;
if (x / mid >= mid) left = mid + 1;
else right = mid;
}
return right - 1;
}
};
这道题还有另一种解法,是利用牛顿迭代法,记得高数中好像讲到过这个方法,是用逼近法求方程根的神器,在这里也可以借用一下,可参见网友Annie Kim's Blog的博客,因为要求x2 = n的解,令f(x)=x2-n,相当于求解f(x)=0的解,可以求出递推式如下:
xi+1=xi - (xi2 - n) / (2xi) = xi - xi / 2 + n / (2xi) = xi / 2 + n / 2xi = (xi + n/xi) / 2
解法二:
class Solution {
public:
int mySqrt(int x) {
if (x == 0) return 0;
double res = 1, pre = 0;
while (abs(res - pre) > 1e-6) {
pre = res;
res = (res + x / res) / 2;
}
return int(res);
}
};
也是牛顿迭代法,写法更加简洁一些,注意为了防止越界,声明为长整型,参见代码如下:
解法三:
class Solution {
public:
int mySqrt(int x) {
long res = x;
while (res * res > x) {
res = (res + x / res) / 2;
}
return res;
}
};
本文参考自:
https://github.com/grandyang/leetcode/ &
https://github.com/azl397985856/leetcode