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我的LeetCode代码仓:https://github.com/617076674/LeetCode
原题链接:https://leetcode-cn.com/problems/climbing-stairs/description/
题目描述:
知识点:动态规划
思路:动态规划
状态定义:f(x) -------- 到达x + 1阶楼梯的方法数量
状态转移:
(1)当x等于0时,f(0) = 1
(2)当x等于1时,f(1) = 2
(3)当x大于等于2时,f(x) = f(x - 1) + f(x - 2)。
时间复杂度和空间复杂度均是O(n)。
JAVA代码:
public class Solution {
public int climbStairs(int n) {
int[] path = new int[n];
if(n == 1 || n == 2){
return n;
}
path[0] = 1;
path[1] = 2;
for (int i = 2; i < n; i++){
path[i] = path[i - 1] + path[i - 2];
}
return path[n - 1];
}
}
LeetCode解题报告: