# Climbing Stairs

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:**Givennwill be a positive integer.

假设梯子有n层，那么如何爬到第n层呢，因为每次只能怕1或2步，那么爬到第n层的方法要么是从第n-1层一步上来的，要不就是从n-2层2步上来的，所以递推公式非常容易的就得出了：dp\[n] = dp\[n-1] + dp\[n-2]。 由于斐波那契额数列的求解可以用递归，所以我最先尝试了递归，拿到OJ上运行，显示Time Limit Exceeded，就是说运行时间超了，因为递归计算了很多分支，效率很低，这里需要用动态规划 (Dynamic Programming) 来提高效率

```
class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        dp = [0] * (n + 1)
        dp[0] = dp[1] = 1
        for x in range(2, n + 1):
            dp[x] = dp[x - 1] + dp[x - 2]
        return dp[n]
```

上述代码的空间复杂度可以化简为O(1)：

```
class Solution:
    # @param {integer} n
    # @return {integer}
    def climbStairs(self, n):
        a = b = 1
        for x in range(2, n + 1):
            a, b = b, a + b
        return b
```


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