zoukankan      html  css  js  c++  java
  • [Algorithm] Fibonacci problem by using Dynamic programming

    vThere are three ways to solve Fibonacci problem

    1. Recursion
    2. Memoize
    3. Bottom-up

    'First Recursion approach:

    def fib(n):
        if n == 1 or n == 2:
            result = 1
        else:
            result = fib(n-1) + fib(n -2)
        
        return result;

    As we can see to calculate fib(5), we need to calculate fib(3) twice and fib(2) three times.

    Time complexity is O(2^n), because for each n from 3, we need to call fib() twice in else block:

    else:
        result = fib(n-1) + fib(n -2)

    To solve the problem, we can use memoize solution to solve repeated calculation.

    deb fib(n, memo):
        if memo[n] != null
            return memo[n]
        if n == 1 or n == 2:
            result = 1
        else:
            result = fib(n - 1) + fib(n-2)
        memo[n] = result
        return result

    Using fib(5) as example: to calulate fib(5) we need to know fib(4)..fib(3), fib(2), fib(1), since we already know fib(1), fib(2) = 1, then we can know fib(3) = 2..fib(4) = 3, fib(5) = 5. 

    Time complexity is O(2n + 1) -> O(n): because we just need to go though memo once. And 2*2 is because of:

    result = fib(n - 1) + fib(n-2)

    We still can improve it by using bottom up approach, because from the previous solution:

    Using fib(5) as example: to calulate fib(5) we need to know fib(4)..fib(3), fib(2), fib(1), since we already know fib(1), fib(2) = 1, then we can know fib(3) = 2..fib(4) = 3, fib(5) = 5. 

    We can clear see the solution the problem by going from bottom (fib(1) & fib(2)) to up (fib(5)):

    def fib_bottom_up(n):
        if n == 1 or n == 2:
            return 1
        bottom_up = new int[n+1]
        bottom_up[1] = 1
        bottom_up[2] = 1
        for i from 3 upto n:
            bottom_up[i] = bottom_up[i-1]+bottom_up[i-2]
    
        return bottom_up[n]

    Time complexity is O(n).


    Notice that some programming language has recursion limit, for example, python has set the limiation to 1000, which mean if you keep calling one function 1000 times, it will throw errors.

    In this sense, bottom up is much better than recursion apporach (recursion and memoize).

  • 相关阅读:
    POJ-2478 Farey Sequence(欧拉函数)
    BZOJ-1103: [POI2007]大都市meg(树状数组)
    NOIP2016模拟 星际争霸(二分)
    HDU-1222 Wolf and Rabbit (欧几里得定理)
    POJ-2689 Prime Distance(线性筛法)
    FZU-2134 上车(树状数组)
    FZU-2236 第十四个目标(树状数组)
    2016年11月12日00:14:27
    FZU-1921 栀子花开(线段树)
    BZOJ3132 上帝造题的七分钟
  • 原文地址:https://www.cnblogs.com/Answer1215/p/10366741.html
Copyright © 2011-2022 走看看