Description

70. Climbing Stairs

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

Answer & Explaining

#include <stdio.h>

int climbStairs(int n) {
    // 特殊情況，若 n 為 1，則只有一種方法
    if (n == 1) return 1;

    int dp1 = 1;  // dp[1]
    int dp2 = 2;  // dp[2]

    // 使用滾動變量計算 dp[i]，避免使用額外陣列
    for (int i = 3; i <= n; i++) {
        int current = dp1 + dp2;
        dp1 = dp2;  // 更新 dp1 為上一階的方法數
        dp2 = current;  // 更新 dp2 為當前階的方法數
    }

    return dp2;  // 最終返回 dp[n]
}

Testing

#include <stdio.h>

int climbStairs(int n) {
    // 特殊情況，若 n 為 1，則只有一種方法
    if (n == 1) return 1;

    int dp1 = 1;  // dp[1]
    int dp2 = 2;  // dp[2]

    // 使用滾動變量計算 dp[i]，避免使用額外陣列
    for (int i = 3; i <= n; i++) {
        int current = dp1 + dp2;
        dp1 = dp2;  // 更新 dp1 為上一階的方法數
        dp2 = current;  // 更新 dp2 為當前階的方法數
    }

    return dp2;  // 最終返回 dp[n]
}

int main() {
    // 範例1
    int n1 = 2;
    printf("Output for n=%d: %d\n", n1, climbStairs(n1));  // 輸出: 2

    // 範例2
    int n2 = 3;
    printf("Output for n=%d: %d\n", n2, climbStairs(n2));  // 輸出: 3

    // 範例3
    int n3 = 5;
    printf("Output for n=%d: %d\n", n3, climbStairs(n3));  // 輸出: 8

    return 0;
}