One, the problem
1. Write a function, enter n to find the nth term of the Fibonacci sequence. The definition of Fibonacci sequence is as follows
F(0) = 0, F(1) = 1
F(N) = F(N-1) + F(N-2), where N> 1.
The Fibonacci sequence starts with 0 and 1, and the subsequent Fibonacci numbers are obtained by adding the previous two numbers.
2. Example
Input: n = 2
Output: 1
Input: n = 5
Output: 5
2. Problem solving method
Problem solving idea:
The definition of Fibonacci sequence is f(n + 1) = f(n) + f(n-1)f(n+1)=f(n)+f(n−1), generating There are several ways to do item nn:
1. Recursion method:
Principle: Split the calculation of f(n)f(n) into two sub-problems f(n-1)f(n−1) and f(n-2)f(n−2), and recurse , With f(0)f(0) and f(1)f(1) as termination conditions.
Disadvantages: A large number of repeated recursive calculations, such as f(n)f(n) and f(n-1)f(n−1). Both downward recursion need to calculate f(n-2)f(n−2) separately Value.
2. Memory recursion method:
Principle: Based on the recursion method, create a new array of length nn to store the digital values from f(0)f(0) to f(n)f(n) during recursion. If a number is repeatedly encountered Use directly from the array, avoiding repeated recursive calculations.
Disadvantages: Memory storage requires additional space of O(N)O(N).
3. Dynamic planning:
Principle: Take the Fibonacci sequence property f(n + 1) = f(n) + f(n-1)f(n+1)=f(n)+f(n−1) as the transfer equation.
From the perspective of computational efficiency and space complexity, dynamic programming is the best solution to this problem.
package com.haoxiansheng.demo01.SwordfingerOffer;
import lombok.extern.slf4j.Slf4j;
import java.util.HashMap;
import java.util.Map;
/**
* @author flame
* @data 2020/10/22
*/
@Slf4j
public class FibDemo {
public static void main(String[] args) {
log.info("fib=>{}", fib(2000));
}
// 为什么选1000000007
//理由大概是“int32类型是十位数，对1e9取模可防止int32溢出”、“1e9+7是质数，
//对质数取模可以尽可能地让模数避免相等”以及“1e9+7是离1e9最近的质数，比较好记”
public static int fib(int n) {
int a =0;
int b = 1;
int sum;
for (int i =0 ; i< n; i++) {
sum = (a + b) % 1000000007;
a = b;
b = sum;
}
return a;
}
int constant = 1000000007;
public int fib1(int n) {
if (n < 2)
return n;
int first = fib(n - 1) % constant;
int second = fib(n - 2) % constant;
return (first + second) % constant;
}
public int fib2(int n) {
return fib(n, new HashMap<>());
}
public int fib(int n, Map<Integer, Integer> map) {
if (n < 2)
return n;
if (map.containsKey(n))
return map.get(n);
int first = fib(n - 1, map) % constant;
map.put(n - 1, first);
int second = fib(n - 2, map) % constant;
map.put(n - 2, second);
int res = (first + second) % constant;
map.put(n, res);
return res;
}
// 非递归解决
public int fib3(int n) {
int constant = 1000000007;
int first = 0;
int second = 1;
while (n-- > 0) {
int temp = first + second;
first = second % constant;
second = temp % constant;
}
return first;
}
}
1000000007Need to look into it. Calculation needs to be understood in the process area