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  • 动态规划系列 Leetcode 70. Climbing Stairs

    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. 1 step + 1 step
    2. 2 steps
    

    Example 2:

    Input: 3
    Output:  3
    Explanation:  There are three ways to climb to the top.
    
    1. 1 step + 1 step + 1 step
    2. 1 step + 2 steps
    3. 2 steps + 1 step

    这个一道基本动态规划题目,做动态规划题目有四个步骤:
    1)确定原问题和子问题
    2)确定状态
    3)确认边界状态的值
    4)确定状态转移方程
     1 #include <stdio.h>
     2 
     3 #include <vector>
     4 class Solution {
     5 public:
     6     int climbStairs(int n) {
     7         std::vector<int> dp(n + 3, 0);
     8         dp[1] = 1;
     9         dp[2] = 2;
    10         for (int i = 3; i <= n; i++){
    11             dp[i] = dp[i-1] + dp[i-2];
    12         }
    13         return dp[n];
    14     }
    15 };
    16 
    17 int main(){
    18     Solution solve;
    19     printf("%d
    ", solve.climbStairs(3));    
    20     return 0;
    21 }

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  • 原文地址:https://www.cnblogs.com/Hwangzhiyoung/p/8733667.html
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