509.Fibonacci-Number.md

509. Fibonacci Number

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题目地址

https://leetcode.com/problems/fibonacci-number/

题目描述

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0,   F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.
Given N, calculate F(N).

Example 1:
Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:
Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:
Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
 
Note:
0 ≤ N ≤ 30.

代码

Approach #1 Recursion

Complexity Analysis

• Time complexity : O(2^N)
• Space complexity : O(N)

class Solution {
  public int fib(int N) {
		if (N <= 1) {
      return N;
    }
    return fib(N - 1) + fib(N - 2);
  }
}

Approach #2 Bottom-Up Approach using memoization

class Solution {
  public int fib(int N) {
    if (N <= 1) {
      return N;
    }
    return memoize(N);
  }
  
  public int memoize(int N) {
    int[] cache = new int[N + 1];
    cache[1] = 1;
    
    for (int i = 2; i <= N; i++) {
      cache[i] = cache[i - 1] + cache[i - 2];
    }
    return cache[N];
  }
}

Approach #3 Top-Down Approach using Memoization

class Solution {
  private Integer[] cache = new Integer[31];
  
  public int fib(int N) {
    if (N <= 1) {
      return N;
    }
    cache[0] = 0;
    cache[1] = 1;
    return memoize(N);
  }
  
  public int memoize(int N) {
    if (cache[N] != null) {
      return cache[N];
    }
    cache[N] = memoize(N-1) + memoize(N-2);
    return memoize(N);
  }
}

Approach #4 Iterative Top-Down Approach

class Solution {
  if (N <= 1) {
    return N;
  }
  if (N == 2) {
    return 1;
  }
  
  int current = 0;
  int prev1 = 1;
  int prev2 = 1;
  
  for (int i = 3; i <= N; i++) {
    current = prev1 + prev2;
    prev2 = prev1;
    prev1 = current;
  }
  
  return current;
}

Approach #5 Matrix Exponetiation

class Solution {
    int fib(int N) {
        if (N <= 1) {
          return N;
        }
        int[][] A = new int[][]{{1, 1}, {1, 0}};
        matrixPower(A, N-1);

        return A[0][0];
    }

    void matrixPower(int[][] A, int N) {
        if (N <= 1) {
          return;
        }
        matrixPower(A, N/2);
        multiply(A, A);

        int[][] B = new int[][]{{1, 1}, {1, 0}};
        if (N%2 != 0) {
            multiply(A, B);
        }
    }

    void multiply(int[][] A, int[][] B) {
        int x = A[0][0] * B[0][0] + A[0][1] * B[1][0];
        int y = A[0][0] * B[0][1] + A[0][1] * B[1][1];
        int z = A[1][0] * B[0][0] + A[1][1] * B[1][0];
        int w = A[1][0] * B[0][1] + A[1][1] * B[1][1];

        A[0][0] = x;
        A[0][1] = y;
        A[1][0] = z;
        A[1][1] = w;
    }
}

Approach #6 Math

class Solution {
    public int fib(int N) {
        double goldenRatio = (1 + Math.sqrt(5)) / 2;
        return (int)Math.round(Math.pow(goldenRatio, N)/ Math.sqrt(5));
    }
}