70 Climbing Stairs

70. Climbing Stairs

1. Question

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.

Example 1:

Input: 2

Output: 2

Explanation:
There are two ways to climb to the top.

1. 1 step + 1 step
2. 2 steps

Example 2:

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

2. Implementation

(1) DP

思路: f(n)表示走到第n个台阶的方式， f(n) = f(n - 1) + f(n - 2), base case是f(1) = 2, f(2) = 2

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }

        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;

        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        } 
        return dp[n];
    }
}

(2) Greedy

思路: 每一步的状态取决于之前的两个状态，所以只需要两个变量存储之前的两个状态即可

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }

        int step1 = 1;
        int step2 = 2;

        for (int i = 3; i <= n; i++) {
            int temp = step2;
            step2 = step1 + step2;
            step1 = temp;
        } 
        return step2;
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(n), 空间复杂度O(n)

Greedy: 时间复杂度O(n), 空间复杂度O(1)