70. Climbing Stairs

描述：
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?

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动态规划

初始状态为
dp[0] = 1, dp [1] = 1
动态规划等式
dp[n] = dp[n-1] + dp[n-2]

不难得到代码

public class Solution {    public int climbStairs(int n) {        if(n < 2) return 1;        int [] dp = new int[3];        dp[0] = 1;        dp[1] = 1;        for(int i = 2; i <= n; i++) {            dp[i % 3] = dp[(i - 1) % 3] + dp[(i - 2) % 3];        }        return dp[n % 3];    }}

其中，对3取模是对空间进行优化，如果去掉的话可以提高运算速度，时间还是空间的抉择看人了

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矩阵求解

可参考XDOJ之feibonaqi的文章，时间复杂度可由动态规划的O(n)下降到O(logn)

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队列求解

另外可用队列求解。代码如下：

public class Solution {    public int climbStairs(int n) {        if (n <= 1) {            return 1;        }        int last = 1, lastlast = 1;        int now = 0;        for (int i = 2; i <= n; i++) {            now = last + lastlast;            lastlast = last;            last = now;        }        return now;    }}