LeetCode70. Climbing Stairs（C++）
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?
Note: Given n will be a positive integer.
Example 1:
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
解题思路：设爬n层楼梯有f(n)种方式，则爬n+1层楼梯有f(n+1)=f(n)+f(n-1);即分为两种情况，一种增加的这一层最后一步是一步爬上去的，该情况有f(n)种，第二种情况最后一层是在倒数第三层走了两步爬上去的，该情况有f(n-1)种。
class Solution {
public:
int climbStairs(int n) {
if(n==1||n==0)
return 1;
int x=1,y=1;
for(int i=2;i<n;i++){
x=x+y;
swap(x,y);
}
return x+y;
}
};
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