LeetCode 70. Climbing Stairs 动态规划问题

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.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 

 is 

n1+ n2

. Because from the 

[n-1]

 point, we can take one single step to reach 

.And from the 

[n-2]

 point, we could take two steps to get there. There is NO overlappingbetween 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.解释的很清楚，这是一个动态规划问题，啥叫动态规划呢？动态规划算法与分治法类似，其基本思想也是将待求解问题分解成若干个子问题，先求解子问题，然后从这些子问题的解得到原问题的解。class Solution {public:int climbStairs(int n) {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;}};