LeetCode 70: 爬樓梯

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.
3. 1 step + 1 step + 1 step
4. 1 step + 2 steps
5. 2 steps + 1 step
這是一個爬樓梯問題，自己18年在面試小米公司的時候遇到過，當時聽完面試官說這道題的時候自己一臉懵逼，面試官後來講完解題思路自己纔有點恍然大悟，昨天在回京的火車上又看到這道題，今天記錄一下。
這個問題實際上跟斐波那契數列非常相似，假設梯子有n層，那麼如何爬到第n層呢，因爲每次只能爬1或2步，那麼爬到第n層的方法要麼是從第 n-1 層一步上來的，要不就是從 n-2 層2步上來的，所以遞推公式非常容易的就得出了：dp[n] = dp[n-1] + dp[n-2]。
Java 解法一：

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

        return dynamicArray[n];
    }

我們可以對空間進行進一步優化，只用兩個整型變量a和b來存儲過程值，首先將 a+b 的值賦給b，然後a賦值爲原來的b，所以應該賦值爲 b-a 即可。這樣就模擬了上面累加的過程，而不用存儲所有的值，參見代碼如下：
Java解法2：

public class Solution {
    public int climbStairs(int n) {
        int a = 1, b = 1;
        while (n-- > 0) {
            b += a; 
            a = b - a;
        }
        return a;
    }
}

或者

 public static int climbStairs4(int n) {
            int a = 1, b = 1;
            while (--n > 0) {
                b += a;
                a = b - a;
            }
            return b;
        }

當然還有使用數組做緩存的方法
Java解法3

public class ClimbingStairs {

    public static int climbStairsWithRecursionMemory(int n) {
        int[] memoryArray = new int[n + 1];
        return subClimbStairsWithRecursionMemory(n - 1, memoryArray) + subClimbStairsWithRecursionMemory(n - 2, memoryArray);

    }

    public static int subClimbStairsWithRecursionMemory(int n, int[] memoryArray) {
        if (n == 1) {
            return 1;
        } else if (n == 2) {
            return 2;
        } else {
            if (memoryArray[n] > 0) {
                return memoryArray[n];
            }
            memoryArray[n] = subClimbStairsWithRecursionMemory(n - 1, memoryArray) + subClimbStairsWithRecursionMemory(n - 2, memoryArray);

            return memoryArray[n];
        }
    }
}

另外還有其他方法
參考文章 算法:Climbing Stairs(爬樓梯) 6種解法
70. Climbing Stairs 爬樓梯問題
https://leetcode.com/problems/climbing-stairs/solution/