刷着Leetcode，就迎来了21岁的生日
the goal of next five years : to be a top coder
（一）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
- 题解参考：https://mio4.github.io/2018/08/27/SwordOnOffer06/
- 可以说是一个典型题目了，最开始直接使用递归做，发现超时，使用迭代优化算法后AC
package DynamicProgramming.ClimbingStairs;
public class Solution {
public static int climbStairs(int n) {
int[] results = {0,1,2,3};
if(n <= 3)
return results[n];
int fibN = 0;
int fibNMinusOne = 3; //第倒数n-1项
int fibNMinusTwo = 2; //第倒数n-2项
for(int i=4; i <= n; i++){
fibN = fibNMinusOne + fibNMinusTwo;
fibNMinusTwo = fibNMinusOne;
fibNMinusOne = fibN;
}
return fibN;
}
public static void main(String[] args){
System.out.println(climbStairs(4)); //5
}
}
- 和斐波那契数列不同的是：边界条件稍微改变（可以从第4项开始迭代）