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?
此题用动太计划解决。
递归式为：dp[n] = dp[n⑴] + dp[n⑵]
爬到第n层，有两种途径，1步从n⑴上来，1下跨两步从n⑵上来。
即要求出爬到第n层的所以方法，需知道爬到第n⑴层，n⑵层的方法。
关于出发点0层，可以定义为有1种方法，即不动。既不跨1步，也不跨两步，就到达。
比0层更低的，定义为0种办法。
这也可看做是Fibonacci求解。
0,1,2,3,5,8,13,21,34,...
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;
}
};
