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LeetCode刷題日記系列 第 28

【Day 28】#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

解析

此題給予一個有n階層的階梯,每次可以走一階或兩階,要求共有幾種走法。

此題是典型的動態規劃(Dynamic Programming)題,可以用Bottom Up(iterative)或Top Down(recursive)兩種DP解法來解。

解法一(Recursion with Memoization)

時間複雜度:O(N)
空間複雜度:O(N)

public class Solution {
    public int climbStairs(int n) {
        int memo[] = new int[n + 1];
        return helper(0, n, memo);
    }
    public int helper(int i, int n, int memo[]) {
        if (i > n) {
            return 0;
        }
        if (i == n) {
            return 1;
        }
        if (memo[i] > 0) {
            return memo[i];
        }
        memo[i] = helper(i + 1, n, memo) + helper(i + 2, n, memo);
        return memo[i];
    }
}

解法二(Dynamic Programming)

時間複雜度:O(N)
空間複雜度:O(1)

public class Solution {
    public int climbStairs(int n) {
        if (n == 1) {
            return 1;
        }
        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) {
            // Every current stage can be achieved by former 1 stage and 2 stages
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}

備註


希望透過記錄解題的過程,可以對於資料結構及演算法等有更深一層的想法。
如有需訂正的地方歡迎告知,若有更好的解法也歡迎留言,謝謝。


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LeetCode刷題日記30

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