[题目]
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?
[分析] 一维动态规划入门题
public class Solution {
// O(n)空间
public int climbStairs0(int n) {
if (n <= 0)
return 0;
if (n == 1)
return 1;
int[] dp = new int[n + 1];
dp[1] = 1;
dp[2] = 2;
for (int i = 3; i <= n; i++)
dp[i] = dp[i - 1] + dp[i - 2]; //递推关系式
return dp[n];
}
// O(1)空间
public int climbStairs(int n) {
if (n <= 0)
return 0;
int[] dp = {1, 1, 1};
for (int i = 2; i <= n; i++) {
dp[2] = dp[1] + dp[0];
dp[0] = dp[1];
dp[1] = dp[2];
}
return dp[2];
}
}
本文介绍使用一维动态规划解决经典爬楼梯问题,通过两种方法优化空间复杂度,详细解释了递推关系式和核心思路。
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