Leetcode:70. Climbing Stairs

Description

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
   这道题的要求是爬n阶楼梯。每次仅仅能够爬1步或2步，总共同拥有多少种不同方法能爬到顶？是动态规划问题。如果要爬到第i阶，能够由i-1和i-2阶1次过去。
   因此dp[i] = dp[i-1] + dp[i-2]。时间复杂度：O(n)，空间复杂度：O(1)

class Solution {
public:
    int climbStairs(int n) {
        int n2 = 0, n1 = 1, res = 0;
        for(int i = 0; i < n; ++ i) {
            res = n2 + n1;
            n2 = n1;
            n1 = res;
        }
        return res;
    }
};

除此之外本题也是求斐波那契数列的第n+1位问题,时间复杂度：O(n)，空间复杂度：O(n),采用动态规划算法复杂度更低

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