70. Climbing Stairs

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

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

Note: Given n will be a positive integer.

public int climbStairs(int n) {    // base cases    if(n <= 0) return 0;    if(n == 1) return 1;    if(n == 2) return 2;        int one_step_before = 2;    int two_steps_before = 1;    int all_ways = 0;        for(int i=2; i<n; i++){    all_ways = one_step_before + two_steps_before;    two_steps_before = one_step_before;        one_step_before = all_ways;    }    return all_ways;}

Here are the steps to get the solution incrementally.

• Base cases:
  if n <= 0, then the number of ways should be zero.
  if n == 1, then there is only way to climb the stair.
  if n == 2, then there are two ways to climb the stairs. One solution is one step by another; the other one is two steps at one time.

• The key intuition to solve the problem is that given a number of stairs n, if we know the number ways to get to the points[n-1] and [n-2] respectively, denoted as n1 andn2 , then the total ways to get to the point [n] is n1 + n2. Because from the [n-1] point, we can take one single step to reach[n]. And from the [n-2] point, we could take two steps to get there. There is NO overlapping between these two solution sets, because we differ in the final step.