LeetCode 70. Climbing Stairs

1. 题目描述

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?

2. 解题思路

非常easy 的一道题目， 可以很容易得到转移方程 ：

f[i]=f[i−1]+f[i−2]

3. code

class Solution {
public:
    int climbStairs(int n) {
        if (n < 3)
            return n;
        int one = 1, two = 2, cur = 0;
        for (int i = 3; i != n + 1; i++){
            cur = one + two;
            one = two;
            two = cur;
        }
        return cur;
    }
};

4. 大神解法

感觉解法和我们的也差不多

/* The problem seems to be a dynamic programming one. Hint: the tag also suggests that! Here are the steps to get the solution incrementally. Base cases: if n <= 0, then the number of ways should be zero. if n == 1, then there is only way to climb the stair. if n == 2, then there are two ways to climb the stairs. One solution is one step by another; the other one is two steps at one time. The key intuition to solve the problem is that given a number of stairs n, if we know the number ways to get to the points [n-1] and [n-2] respectively, denoted as n1 and n2 , then the total ways to get to the point [n] is n1 + n2. Because from the [n-1] point, we can take one single step to reach [n]. And from the [n-2] point, we could take two steps to get there. There is NO overlapping between these two solution sets, because we differ in the final step. Now given the above intuition, one can construct an array where each node stores the solution for each number n. Or if we look at it closer, it is clear that this is basically a fibonacci number, with the starting numbers as 1 and 2, instead of 1 and 1. The implementation in Java as follows: */
public int climbStairs(int n) {
    // base cases
    if(n <= 0) return 0;
    if(n == 1) return 1;
    if(n == 2) return 2;

    int one_step_before = 2;
    int two_steps_before = 1;
    int all_ways = 0;

    for(int i=2; i<n; i++){
        all_ways = one_step_before + two_steps_before;
        two_steps_before = one_step_before;
        one_step_before = all_ways;
    }
    return all_ways;
}