70.Climbing Stairs(Dp入门)
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2022-03-15 20:35:35
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参考博客:https://www.cnblogs.com/lightwindy/p/8476881.html
题目:
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.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
- 1 step + 1 step
- 2 steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
- 1 step + 1 step + 1 step
- 1 step + 2 steps
- 2 steps + 1 step
解析:动态规划的简单案例。
state:dp[i]表示爬到第i个楼梯的所有方法的和
function:dp[i]=dp[i-1]+dp[i-2],每次走一步或者两步,所以dp[i]就是它前一步和前两步方法的和
initicl:dp[0]=1,dp[1]=2,分别为n=1和n=2的结果
end:return dp[n-1]
方法:
class Solution {
public:
int climbStairs(int n) {
if(n<=1) return 1;
//tips:若此处vector初始化时没有指定元素数量(容量),则只能使用pusn_back向容器中添加元素,不能使用元素的下标直接赋值。
//vector的下标运算符用于访问已存在的元素,不能用于赋值(即在用于初始化之前必须指定容器的容量大小)
vector<int> dp(n);
dp[0] = 1;
dp[1] = 2;
for(int i=2;i<n;i++)
dp[i]=dp[i-1]+dp[i-2];
return dp.back();//返回vector中尾元素的引用
}
};