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Software Development

## 【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
``````

## 解法一(Recursion with Memoization)

``````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)

``````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];
}
}
``````

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