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算法笔记: 力扣#70 爬楼梯

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

class :
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n < 3:
            return n
        first, second = 1, 2
        for _ in range(3, n+1):
            first, second = second, first + second
        return second

Java 实现

class  {
    public int climbStairs(int n) {
        if(n < 3){ return n; }
        int first = 1;
        int second = 2;
        for(int i = 3; i <= n; i++){
            int temp = first + second;
            first = second;
            second = temp;
        }
        return second;
    }
}

时间复杂度

O(n).

空间复杂度

O(1).

链接

70. Climbing Stairs
70. 爬楼梯
 (English version) Algorithm Notes: Leetcode#70 Climbing Stairs