Problem
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
Solution (Java)
class Solution {
public int climbStairs(int n) {
if (n == 0 || n == 1 || n == 2) return n;
int[] mem = new int[n];
mem[0] = 1;
mem[1] = 2;
for (int i = 2; i < n; i ++) {
mem[i] = mem[i - 1] + mem[i - 2];
}
return mem[n - 1];
}
}
Solution (Javascript)
/**
* @param {number} n
* @return {number}
*/
var climbStairs = function(n) {
var dp = [0, 1];
for (var i = 0; i < n; i++) {
dp = [dp[1], dp[0] + dp[1]];
}
return dp[1];
};
Explain:
f(x) = f(x - 1) + f(x - 2)。这里用了滚动数组，只保留前两个值，注意处理 n = 0, 1, 2 时的特殊情况。
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).