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.

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

解题要点：

使用dp，累计加加加，返回最后一个值。

class Solution {
    public int climbStairs(int n) {
        if(n == 1) return 1;
        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;
        for(int i = 3; i <= n; i++)
            dp[i] = dp[i-2] + dp[i-1];
                
        return dp[n];
    }
}

Complexity Analysis

• Time complexity : O(n). Single loop upto n.

• Space complexity : O(n). dp array of size n is used.

学习解法：

使用斐波那契数列（Fibonacci sequence），返回最后一个值。

public class Solution {
    public int climbStairs(int n) {
        if (n == 1) {
            return 1;
        }
        int first = 1;
        int second = 2;
        for (int i = 3; i <= n; i++) {
            int third = first + second;
            first = second;
            second = third;
        }
        return second;
    }
}

Complexity Analysis

• Time complexity : O(n). Single loop upto n is required to calculate n^(th）Fibonacci number.

• Space complexity : O(1). Constant space is used.