算法 — easy —爬楼梯（dp[i] = dp[i-1] + dp[i-2]）

一开始想到的是分类列举，那可太麻烦了，但是看到评论区的hxd居然真的有列举法的，给出的理由是当台阶数到达46，会出现溢出；所以列举出了45种情况哈哈哈哈

列举方法：

public int climbStairs(int n) {
    
    int result = 0;
    
    switch(n){
    case 1: result = 1; break;
    case 2: result = 2; break;
    case 3: result = 3; break;
    case 4: result = 5; break;
    case 5: result = 8; break;
    case 6: result = 13; break;
    case 7: result = 21; break;
    case 8: result = 34; break;
    case 9: result = 55; break;
    case 10: result = 89; break;
    case 11: result = 144; break;
    case 12: result = 233; break;
    case 13: result = 377; break;
    case 14: result = 610; break;
    case 15: result = 987; break;
    case 16: result = 1597; break;
    case 17: result = 2584; break;
    case 18: result = 4181; break;
    case 19: result = 6765; break;
    case 20: result = 10946; break;
    case 21: result = 17711; break;
    case 22: result = 28657; break;
    case 23: result = 46368; break;
    case 24: result = 75025; break;
    case 25: result = 121393; break;
    case 26: result = 196418; break;
    case 27: result = 317811; break;
    case 28: result = 514229; break;
    case 29: result = 832040; break;
    case 30: result = 1346269; break;
    case 31: result = 2178309; break;
    case 32: result = 3524578; break;
    case 33: result = 5702887; break;
    case 34: result = 9227465; break;
    case 35: result = 14930352; break;
    case 36: result = 24157817; break;
    case 37: result = 39088169; break;
    case 38: result = 63245986; break;
    case 39: result = 102334155; break;
    case 40: result = 165580141; break;
    case 41: result = 267914296; break;
    case 42: result = 433494437; break;
    case 43: result = 701408733; break;
    case 44: result = 1134903170; break;
    case 45: result = 1836311903; break;
    
    }
    return result;
}

分析：观察上面给出的答案，可以得出结论dp[i] = dp[i-1] + dp[i-2]。斐波那契的一般思路是从后往前看，比如想求n=7时的解，根据公式dp[i] = dp[i-1] + dp[i-2]很容易就想到n=7时等于n=6时+n=5时，n=6和5又有n=5 + n=4和n=4 + n=3，这么一步步下去就是递归了。 但"Java"的做法是从前往后看的，第一位 i1=1，第二位 i2=2，那第三位呢，就是temp = i1+i2，然后让i2变成新的i1，就把i2赋值给i1，i2自己则变成下一位即temp，这样一步一步往后移，用while循环控制移动几次，最后输出结果就可以。

代码：

class Solution {
    public int climbStairs(int n) {
        if(n <= 3)
            return n;
        int dp1 = 1;
        int dp2 = 2;
        int i = 2;
        int temp = 0;
        while(i < n)
        {
            temp = dp1 + dp2;
            dp1 = dp2;
            dp2 = temp;
            i++;
        }
        return temp;
    }
}