70.Climbing Stairs

You are climbing a stair case. It takesnsteps 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.

Analysis:

https://leetcode.com/problems/climbing-stairs/discuss/25299/Basically-it's-a-fibonacci.

The problem seems to be a_dynamic programming_one.Hint: the tag also suggests that!
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 asn1andn2, then the total ways to get to the point[n]isn1 + 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.

Now given the above intuition, one can construct an array where each node stores the solution for each number n. Or if we look at it closer, it is clear that this is basically a fibonacci number, with the starting numbers as 1 and 2, instead of 1 and 1.

Solution

So the"all_ways"corresponds to the number of solutions to get to the point[n].
And"one_step_before"refers to the number of solutions until the point[n-1],
"two_steps_before"refers to the number of solution until the point[n-2].

From the point[n-1], we take one step to reach the point[n].
From the point[n-2], we take a two-steps leap to reach the point[n].

So it goes without saying that the total number of solution to reach the point[n]should be[n-1] + [n-2].

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 = 3; 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;
}