- 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?
解题思路:
动态规划 DP(Dynamic Programming) 入门题
state: dp[i] 表示爬到第 i 个楼梯的所有方法的和
- function: dp[i] = dp[i-1] + dp[i-2] // 因为每次走一步或者两步, 所以 f[i] 的方法就是它一步前和两步前方法加和
- initial: dp[0] = 0; dp[1] = 1
- end : return f[n]
- Java: Method 1: Time: O(n), Space: O(n)
- public int climbStairs(int n) {
- int[] dp = new int[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];
- }
- Java: Method 2: Time: O(n), Space: O(1)
- public int climbStairs(int n) {
- if (n == 0 || n == 1 || n == 2){
- return n;
- }
- int [] dp = new int[3];
- dp[1] = 1;
- dp[2] = 2;
- for (int i =3; i <= n; i++) {
- dp[i%3] = dp[(i-1)%3] + dp[(i-2)%3];
- }
- return dp[n%3];
- }
- Java: Method 3: Time: O(n), Space: O(1)
- public class Solution {
- public int climbStairs(int n) {
- int[] dp = new int[]{0,1,2};
- if(n < 3) return dp[n];
- for(int i = 2; i < n; i++){
- dp[0] = dp[1];
- dp[1] = dp[2];
- dp[2] = dp[0] + dp[1];
- }
- return dp[2];
- }
- }
- Python:
- class Solution:
- # DP Time: O(n) Space: O(1)
- def climbStairs(self, n):
- prev, current = 0, 1
- for i in xrange(n):
- prev, current = current, prev + current,
- return current
- # Recursion Time: O(2^n) Space: O(n)
- def climbStairs1(self, n):
- if n == 1:
- return 1
- if n == 2:
- return 2
- return self.climbStairs(n - 1) + self.climbStairs(n - 2)
来源: http://www.bubuko.com/infodetail-2508554.html