Climbing Stairs，climbingstairs

Climbing Stairs，climbingstairs

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?

此題用動太規劃解決。

遞迴式為：dp[n] = dp[n-1] + dp[n-2]

爬到第n層，有兩種途徑，一步從n-1上來，一下跨兩步從n-2上來。

即要求出爬到第n層的所以方法，需知道爬到第n-1層，n-2層的方法。

關於起點0層，可以定義為有一種方法，即不動。既不跨一步，也不跨兩步，就達到。

比0層更低的，定義為0種辦法。

這也可看作是Fibonacci求解。

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

class Solution {public:    int climbStairs(int n) {        if (n == 0 || n == 1)            return 1;                int stepOne = 1, stepTwo = 1;        int allWays;        for (int i=2; i<=n; i++) {            allWays = stepOne + stepTwo;            stepTwo = stepOne;            stepOne = allWays;        }                return allWays;    }};