70. Climbing Stairs
- Hardness: \(\color{green}\textsf{Easy}\)
- Ralated Topics:
Math、Dynamic Programming、Memoization
一、題目
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 step + 1 step
- 2 steps
Example 2:
- Input: n = 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
Constraints:
1 <= n <= 45
二、分析
- 經典的動態規劃問題，因為只有
1-step與2-steps兩個選擇，所以n階的樓梯的走法會是n-1階與n-2階走法的總和。 - 令
n階走法的數目為dp[n]，dp[n] = dp[n-1] + dp[n-2]。 - 注意初始條件
dp[0] = 1，dp[1] = 1。 - 可以發現此為費式數列，即
1,1,2,3,5,8,13,21...。
三、解題
1. DP
- Time complexity: \(O()\)
- Space complexity: \(O()\)
int climbStairs(int n) {
int dp[n+1];
dp[0] = 1, dp[1] = 1;
for (int i = 2; i <= n; i++) {
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
}