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70. Climbing Stairs

70. Climbing Stairs

作者: 衣介书生 | 来源:发表于2018-04-13 18:58 被阅读10次

    题目分析

    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
    

    代码一

    时间复杂度为O(2^n),Time Limit Exceeded。

    class Solution {
        public int climbStairs(int n) {
            if(n <= 2) {
                return n;
            }
            return climbStairs(n - 2) + climbStairs(n - 1);
        }
    }
    

    代码二

    动态规划的解法

    class Solution {
        public int climbStairs(int n) {
            if (n == 1) return 1;
            int[] num = new int[n + 1];
            num[0] = 1;
            num[1] = 1;
            for (int i = 2; i <= n; i++) {
                num[i] = num[i - 1] + num[i - 2];
            }
            return num[n];
        }
    }
    

    代码三

    class Solution {
        public int climbStairs(int n) {
            if (n == 1) return 1;
            int t1 = 1, t2 = 1, res = 0;
            for(int i = 2; i <= n; i++) {
                res = t1 + t2;
                t2 = t1;
                t1 = res;
            }
            return res;
        }
    }
    

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