Leetcode 70. Climbing Stairs-每次能走1步或2步臺階,輸入n,求總的方法數
阿新 • • 發佈:2019-01-06
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
public class Leetcode_70_ClimbingStairs { public static void main(String[] args) { Leetcode_70_ClimbingStairs leetcode_70_climbingStairs = new Leetcode_70_ClimbingStairs(); System.out.println(leetcode_70_climbingStairs.climbStairs(5));//8 } // 1 1 // 2 2 // 3 3 // 4 5 // 5 8 // 8=5+3 public int climbStairs2(int n) { if (n == 1 || n == 2) { return n; } int a = 1; int b = 2; int c = 0; for (int i = 3; i <= n; i++) { c = a + b; a = b; b = c; } return c; } /** * 優化之後 * @param n * @return */ public int climbStairs(int n) { int a = 0; int b = 1; int c = 0; for (int i = 1; i <= n; i++) { c = a + b; a = b; b = c; } return c; } }
Runtime: 3 ms, faster than 41.36% of Java online submissions for Climbing Stairs.
方法2:
/** * 分治法一般使用遞迴 * * @param n * @return */ //Time Limit Exceeded public int climbStairs4(int n) { if (n == 1) { return 1; } if (n == 2) { return 2; } return climbStairs(n - 1) + climbStairs(n - 2); }
Time Limit Exceeded