Leetcode-70. Climbing Stairs
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好久没有刷题了,感觉手都生了。今天来道简单点的题目——爬楼梯;
题目如下:
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?
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看到题目我们第一反应就是利用递归的思想,所以我的第一个版本的代码如下:
class Solution {public:int climbStairs(int n) {if (n == 0)return 0;else if (n == 1)return 1;else if (n == 2)return 2;else return climbStairs(n-1) + climbStairs(n-2);}};
这种情况下当n=4的时候运行超时,很明显,我们可以用空间来换时间。每当我们得到一个n的结果的时候就将它保存到容器re中,当下次需要用到某个n的时
候可以直接从容器中取,而无需再利用递归再求一次值。递归求解方程式climbStairs(n)=climbStairs(n-1) + climbStairs(n-2);其实也是动态规划的思想;改进
之后的代码如下:
class Solution {public:int climbStairs(int n) {vector<int>re(n+1,0);if (n == 0)return 0;else if (n == 1)return 1;else if (n == 2)return 2;else {re[1] = 1;re[2] = 2;for (int i = 3; i <= n; i++){re[i] = re[i - 1] + re[i - 2];}return re[n];}}};运行结果:
Submission Result: Accepted More Details
Next challenges: (H) Wildcard Matching (H) Palindrome Partitioning II (M) Android Unlock Patterns
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