花花酱 LeetCode. 69 Sqrt(x)

zxi — Wed, 17 Jan 2018 02:41:12 +0000

题目大意：让你实现开根号函数，只需要返回整数部分。

Problem:

Implement int sqrt(int x).

Compute and return the square root of x.

x is 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...

Solution 1: Brute force

Time complexity: sqrt(x)

C++

// Author: Huahua
// Running time: 40 ms
class Solution {
public:
  int mySqrt(int x) {
    if (x <= 1) return x;
    for (long long s = 1; s <= x; ++s)
      if (s * s > x) return s - 1;
    return -1;
  }
};

C++ div

// Author: Huahua
// Running time: 162 ms
class Solution {
public:
  int mySqrt(int x) {
    if (x <= 1) return x;
    for (int s = 1; s <= x; ++s)
      if (s > x / s) return s - 1;
    return -1;
  }
};

Java

// Author: Huahua
// Running time: 71 ms
class Solution {
  public int mySqrt(int x) {
    if (x <= 1) return x;
    for (long s = 1; s <= x; ++s)
      if (s * s > x) return (int)s - 1;
    return -1;
  }
}

Python3 TLE

"""
Author: Huahua
TLE: 602 / 1017 test cases passed.
"""
class Solution:
  def mySqrt(self, x):
    if x <= 1: return x
    s = 1
    while True:
      if s*s > x: return s - 1
      s += 1
    return -1

Solution 2: Binary search

Time complexity: O(logn)

C++

// Author: Huahua
// Running time: 12 ms
class Solution {
public:
  int mySqrt(int x) {      
    long l = 1;
    long r = static_cast(x) + 1;
    while (l < r) {
      long m = l + (r - l) / 2;
      if (m * m > x) { 
        r = m;
      } else {
        l = m + 1;
      }
    }
    // l: smallest number such that l * l > x
    return l - 1;
  }
};

Java

// Author: Huahua
// Running time: 47 ms
class Solution {
  public int mySqrt(int x) {      
    int l = 1;
    int r = x;
    while (l <= r) {
      int m = l + (r - l) / 2;
      if (m > x / m) {
        r = m - 1;
      } else {
        l = m + 1;
      }
    }
    return r;
  }
}

Python3

"""
Author: Huahua
Running time: 119 ms
"""
class Solution:
  def mySqrt(self, a):
    l = 1
    r = a
    while l <= r:
      m = l + (r - l) // 2
      if m * m > a:
        r = m - 1
      else:
        l = m + 1
    return r;

Solution 3: Newton’s method

C++ / float

// Author: Huahua
// Running time: 28 ms
class Solution {
public:
    int mySqrt(int a) {
      constexpr double epsilon = 1e-2;
      double x = a;
      while (x * x - a > epsilon) {
        x = (x + a / x) / 2.0;
      }
      return x;
    }
};

C++ / int

// Author: Huahua
// Running time: 27 ms
class Solution {
public:
  int mySqrt(int a) {
    // f(x) = x^2 - a, find root of f(x)
    // Newton's method
    // f'(x) = 2x
    // x' = x - f(x) / f'(x) = x - (1/2*x - a/(2*x))
    //    = (x + a / x) / 2
    int x = a;
    while (x > a / x)
      x = (x + a / x) / 2;
    return x;
  }
};

Java

// Author: Huahua
// Running time: 44 ms
class Solution {
  public int mySqrt(int a) {    
    long x = a;
    while (x * x > a)
      x = (x + a / x) / 2;    
    return (int)x;
  }
}

Python3

"""
Author: Huahua
Running time: 100 ms
"""
class Solution:
  def mySqrt(self, a):
    x = a
    while x * x > a:
      x = (x + a // x) // 2
    return x

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花花酱 LeetCode. 69 Sqrt(x)

Solution 1: Brute force

C++

C++ div

Java

Python3 TLE

Solution 2: Binary search

C++

Java

Python3

Solution 3: Newton’s method

C++ / float

C++ / int

Java

Python3