70.Climbing Stairs

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Dynamic Programming, Easy

Question

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.

Example 1:

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Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

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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

My Answer

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class Solution {
    public int climbStairs(int n){
        if(n==1)
            return 1;
        int[] res = new int[n];
        res[0] = 1;
        res[1] = 2;
        for(int i=2;i<res.length;i++)
            res[i] = res[i-1]+res[i-2];
        return res[n-1];
    }
}

With less space:

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

Running time: O(n), 4ms

Understanding

There are two ways:

Based on the res[n-1], u just need to climb one stair once

Based on the res[n-2], u just need to climb two stairs once

So it can be concluded to a Fibonacci formula