Leetcode ☞ 70. Climbing Stairs
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Total Accepted: 96079 Total Submissions: 265268 Difficulty: Easy
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?
我的AC(0ms,最快):
<span style="font-size:18px;">int climbStairs(int n) { int ans[n]; ans[0] = 0, ans[1] = 1, ans[2] = 2; for(int i = 3; i <= n ; i++){ ans[i] = ans[i-1] + ans[i - 2]; } return ans[n];}</span>迭代比递归好!
分析:
此题在北邮机试PDF里有,此前已详细看过。动态规划部分。
思路是分治:
n的上一状态只有两种情况:1、从n-1跳一层变成n。2、从n-2跳两层变成n 【注意! 不考虑n-2跳两次的情况,因为这不是上一状态,而是上两个状态了!】
每种情况下的走法都只有1种。
因此ans[n] = ans[n-1] + ans[n-2]。
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