LeetCode 子 Climbing Stairs
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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?
这是一个很简单的动态规划问题:假设S(n)为n个台阶时的# ways,我们可以推出S(n)=S(n-1)+S(n-2)。关键是如何实现,本来我想到的是递归的求发现时间太久了,大致估计了下T(n)=O(n*n)。后来改用一个数组存储0到n的Steps.具体的实现如下:
class Solution {public: int climbStairs(int n) { if(n==0||n==1||n==2){ return n; } int A[n]; A[0]=1;A[1]=1;A[2]=2; for (int temp=3;temp<=n;temp++){ A[temp]=A[temp-1]+A[temp-2]; } return A[n]; }};所以说选着什么数据结构和算法很是关键,如果递归不能实现问题的变小,最好是想想其他办法。
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