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?
Example 1:
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| Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
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Example 2:
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| 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
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Constraints:
动态规划
很容易发现, 记f(n)为上第n层楼梯的方法数量, 则有f(n) = f(n-1) + f(n-2). f(0) = 1, f(1) = 1.
动态规划出f(n)即可
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| class Solution {
public int climbStairs(int n) {
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for(int i = 2; i < n + 1; ++i)
dp[i] = dp[i-1] + dp[i-2];
return dp[n];
}
}
|