Algorithm complexity O (logn) Comments

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• A .O (logn) code for small to prove
• II. Typical time complexity
• III. Common L O G N logN Algorithm
  • 1. binary search
  • 2. Euclidean algorithm
  • 3. exponentiation
• IV. $$ library log function
• Finally, and most important basic

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A .O (logn) code for small to prove

Let's look at the following piece of code:

int cnt = 1;

while (cnt < n)
{
    cnt *= 2;
    //时间复杂度为O(1)的程序步骤序列
}

Since cnt after multiplying by 2 each are more close to n, that is, there after x times, thereby cnt will be out of the loop is greater than n, the 2  x  = n = n 2x, i.e. x = L O G  2  nx = log2n, so the complexity of this loop is O (logn)

II. Typical time complexity

$c$ 常数
$logN$ 对数级
$log ^ 2N$ 对数平方根
$N$ 线性级
$NlogN$
$N ^ 2$ 平方级
$N ^ 3$ 立方级
$2 ^ N$ 指数级

Thus we can see that, L O G algorithm's efficiency is the highest N logN

III. Common L O G N logN Algorithm

 

1. binary search

- (int)BinarySearch:(NSArray *)originArray element:(int)element
{
    int low, mid, high;
    low = 0; high = (int)originArray.count - 1;
    while (low <= high) {
        mid = (low + high) / 2;
        if ([originArray[mid] intValue] < element) {
            low = mid + 1;
        } else if ([originArray[mid] intValue] > element) {
            high = mid -1;
        } else {
            return mid;
        }
    }
    
    return -1;
}

 

2. Euclidean algorithm

- (unsigned int)Gcd:(unsigned int)m n:(unsigned int)n
{
    unsigned int Rem;
    while (n > 0) {
        Rem = m % n;
        m = n;
        n = Rem;
    }
    return m;
}

 

3. exponentiation

- (long)Pow:(long)x n:(unsigned int)n
{
    if (n == 0) {
        return 1;
    }
    if (n == 1) {
        return x;
    }
    
    if ([self isEven:n]) {
        return [self Pow:x * x n:n / 2];
    } else {
        return [self Pow:x * x n:n / 2] * x;
    }
}

- (BOOL)isEven:(unsigned int)n
{
    if (n % 2 == 0) {
        return YES;
    } else {
        return NO;
    }
}

IV. $$ library log function

In $$ Curry log () function and log2 () function

log () function in base default base e of natural logarithm

Base number log2 () function is apparently slightly qwq 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl

int main()
{
    cout << log(M_E) << endl;
    cout << log2(2) << endl;
    return 0;
}

Then we get

1
1

the result of

$$ Curry and two constants M_E M_PI
M_E represents the natural logarithm base number E
M_PI π represents the circle ratio

Finally, and most important basic

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