70. Climbing Stairs dynamic programming

本文主要是介绍70. Climbing Stairs dynamic programming，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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.

首先是递归的算法，但是会超时

int climbStairs(int n) {if (n <= 0)return 0;else if (n == 1)return 1;else if (n == 2)return 2;return climbStairs(n - 1) + climbStairs(n - 2);
}

接着用一个数组存储到第i步有多少种走法，但是这样的空间复杂度为O(n)；

int climbStairs(int n) {vector<int>my;if (n <= 0)return 0;else if (n == 1)return 1;else if (n == 2)return 2;my.push_back(0);my.push_back(1);my.push_back(2);for (int i = 3; i <= n; i++){my.push_back(my[i - 1] + my[i - 2]);}return my[n];
}

最后，将空间复杂度降为O（1）；

int climbStairs(int n) {if (n == 0)return 0;if (n == 1)return 1;if (n == 2)return 2;int pre = 1, cur = 2, next = 0;for (int i = 3; i <= n; i++){next = pre + cur;pre = cur;cur = next;}return cur;
}

这篇关于70. Climbing Stairs dynamic programming的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---