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leetCode 70. Climbing Stairs

时间:2017-04-04 00:36:09      阅读:193      评论:0      收藏:0      [点我收藏+]

题目如下:

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.

思路:

当n=1,结果为1

当n=2,结果为2

当n=3,结果为3

当n=4,结果为7

易递推[n]=[n-1]+[n-2],是一个斐波纳切数列。

 

本题代码:

int climbStairs(int n) {
    if (n<=0) return 0;
    if (n==1) return 1;
    if (n==2) return 2;
    
    int one_step_before = 2;
    int two_steps_before = 1;
    int all_ways = 0;
    
    for(int i=3; i<=n; i++) {
        all_ways = one_step_before + two_steps_before;
        two_steps_before = one_step_before;
        one_step_before = all_ways;
    }
    return all_ways;
}

 

leetCode 70. Climbing Stairs

原文:http://www.cnblogs.com/pinganzi/p/6664135.html

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