Climbing Stairs
2013-11-05 10:24
169 查看
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?
思路:当前第n步等于前n-1步走法 + 前n-2步走法,即f(n) = f(n-1) + f(n-2)
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
思路:当前第n步等于前n-1步走法 + 前n-2步走法,即f(n) = f(n-1) + f(n-2)
class Solution { public: int climbStairs(int n) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if(n <= 1) return 1; int f[n+1]; f[0] = 1; f[1] = 1; for(int i=2; i<=n; i++){ f[i] = f[i-1] + f[i-2]; } return f ; } };
相关文章推荐
- [LeetCode] Climbing Stairs
- Climbing Stairs
- leetcode_question_70 Climbing Stairs
- LeetCode 之 Climbing Stairs (方法2)
- Climbing Stairs
- Climbing Stairs
- [leetcode]Climbing Stairs
- [leetcode]_Climbing Stairs
- 52_leetcode_Climbing Stairs
- 【九度】题目1388:跳台阶 && 【LeetCode】Climbing Stairs
- Climbing Stairs
- leetcode 虐我篇之(二十一)Climbing Stairs
- Climbing Stairs 爬楼梯问题,每次可以走1或2步,爬上n层楼梯总方法 (变相fibonacci)
- leetcode:Climbing Stairs
- [LeetCode] Climbing Stairs
- Climbing Stairs
- Climbing Stairs
- LeetCode – Refresh – Climbing Stairs
- [leetcode] #70 Climbing Stairs
- LeetCode-70 Climbing Stairs(斐波那契数列)