Climbing Stairs

• LeetCode: Climbing Stairs

• Problem:

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.

• Examples:

Input: 2
Output:  2
Explanation:  There are two ways to climb to the top.

1. 1 step + 1 step
2. 2 steps

Input: 3
Output:  3
Explanation:  There are three ways to climb to the top.

1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

• Analysis:

F(n) is the number of the distinct ways to climb to the top n.

F(n) = F(n - 2) + F(n - 1), n is from n - 2 steps and n - 1 steps

It's simlar to Fibnacci

• Code - Java:

public class Solution {
    public int climbStairs(int n) {
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 1;
        }
        int a = 1;
        int b = 1;
        int c = 0;
        for (int i = 2; i <= n; i++) {
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }
}

• Code - C++:

class Solution {
public:
    int climbStairs(int n) {
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 1;
        }
        int a = 1;
        int b = 1;
        int c = 0;
        for (int i = 2; i <= n; i++) {
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }
};

• Code - Python:

class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0:
            return 1
        if n == 1:
            return 1
        a, b, c = 1, 1, 0
        for i in range(2, n + 1):
            c = a + b
            a = b
            b = c
        return c

• Code - C:

int climbStairs(int n) {
    if (n == 0) {
        return 1;
    }
    if (n == 1) {
        return 1;
    }
    int a = 1;
    int b = 1;
    int c = 0;
    for (int i = 2; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return c;
}

• Time Complexity: O(N), N is the input n

• Space Complexity: O(1)