## 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.

## 代码

``````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;
}
};

``````

xi+1=xi - (xi2 - n) / (2xi) = xi - xi / 2 + n / (2xi) = xi / 2 + n / 2xi = (xi + n/xi) / 2

``````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);
}
};

``````

``````class Solution {
public:
int mySqrt(int x) {
long res = x;
while (res * res > x) {
res = (res + x / res) / 2;
}
return res;
}
};
``````

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