[数据结构与算法]LeetCode-70. Climbing Stairs [C++][Java]

LeetCode-70. Climbing Stairshttps://leetcode.com/problems/climbing-stairs/

题目描述

You are climbing a staircase. It takes?n?steps to reach the top.

Each time you can either climb?1?or?2?steps. In how many distinct ways can you climb to the top?

Example 1:

Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: n = 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

Constraints:

• 1 <= n <= 45

解题思路

【C++】

1. 动态规划

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 2) {return n;}
        vector<int> dp(n + 1, 1);
        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
};

2. 空间压缩的动态规划

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 2) {return n;}
        int pre2 = 1, pre1 = 2, cur;
        for (int i = 3; i <= n; i++) {
            cur = pre1 + pre2;
            pre2 = pre1;
            pre1 = cur;
        }
        return cur;
    }
};

【Java】

1. 动态规划

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {return n;}
        int[] dp = new int[n + 1];
        Arrays.fill(dp, 1);
        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
}

2. 压缩空间的动态规划

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {return n;}
        int pre2 = 1, pre1 = 2, cur = 0; //必须初始化
        for (int i = 3; i <= n; i++) {
            cur = pre1 + pre2;
            pre2 = pre1;
            pre1 = cur;
        }
        return cur;
    }
}