Lintcode Climbing Stairs

You is climbing a stair case. It takes n steps to reach the top.

Each time you can either climb 1 or 2 steps. In what many distinct ways can you climb to the top?

Example

Given An example n=3, 1+1+1=2+1=1+2=3

Return 3

For the problem, try-to-think about it on this.

Firstly, we define the problem of DP (n) as the ways of approaching to n stairs.

The problem of DP (n) depends on DP (N-1) and DP (N-2). Then the DP (n) = DP (N-1) + DP (n-2). Because there is a possibilities for DP (n) happen due to the rule that either it's accomplished by step 1 or Step 2 STA IRS before approaching to N.

Initialize the DP (0) =1 and DP (1) = 1.

Solve problem DP (n)

1  Public classSolution {2     /**3      * @paramN:an integer4      * @return: An integer5      */6      Public intClimbstairs (intN) {7         //Write your code here8         if(N <= 1) {9             return1;Ten         } One         intLast = 1, lastlast = 1; A         intnow = 0; -          for(inti = 1; I < n; i++) { -now = last +Lastlast; theLastlast =Last ; -Last =Now ; -         } -         returnNow ; +     } -}

Then this problem is a Fibonacci sequence

Lintcode Climbing Stairs