Leetcode 70. Climbing Stairs

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
解析：
斐波那契数列问题。n=0时，结果为0；n=1时，结果为1；n=2时，可以两步每一步1个台阶，也可以一步。当台阶数为n时，可以从n-1处一步台阶过来，也可以从n-2处两步台阶过来。因此当前的结果只与前两个结果有关。
解法1:动态规划

class Solution {
public:
    int climbStairs(int n) {
        vector<int> dp(n, 0);
        if(n==1)
            return 1;
        if(n==2)
            return 2;
        dp[0] = 1;
        dp[1] = 2;
        for(int i=2; i < n; i++)
            dp[i] = dp[i-1] + dp[i-2];
        return dp[n-1];
    }
};

运行结果：
解法2:
由上面分析可知道，当前结果只与前两个结果相关，因此定义两个变量，存放前两个的数值。具体代码如下：

class Solution {
public:
    int climbStairs(int n) {
        int prev = 0;
        int cur = 1;
        for(int i=1; i<=n; ++i)
        {
            int temp = cur;
            cur += prev;
            prev = temp;
        }
        return cur;
    }
};

运行结果：