Leetcode: climbing_stairs

I. Question

There are n steps to climb the stairs. Each time you can climb the level 1 or Level 2, how many crawling methods do you climb to the top?

Ii. Analysis

If F (n) is set, it indicates the number of different methods used to climb the n-level stair. to climb the n-level stair, there are two options:

1. Step forward from n-1

2. Two steps forward from The N-2 order

So there is, F (n) = f (n-1) + f (n-2) Isn't that the Fibonacci series?

Therefore,

Method 1: Iteration

Method 2: Recursion

Method 3: formula F (n) = (1/√ 5) * {[(1 + √ 5)/2] ^ N-[(1-√ 5) /2] ^ n}

Class solution {public: int climbstairs (int n) {int flag; int stair0 = 1; int stair1 = 1; if (n <= 0) return 0; if (n = 1) return 1; for (INT I = 2; I <= N; I ++) {flag = stair1; stair1 = stair0 + stair1; stair0 = flag;} return stair1 ;}}; formula: Class solution {public: int climbstairs (int n) {double flag = SQRT (5 ); return floor (POW (1 + flag)/2, n + 1) + POW (1-flag)/2, n + 1 )) /flag + 0.5 );}};

Leetcode: climbing_stairs