[LeetCode]Climbing Stairs
来源:互联网 发布:炒股软件免费下载 编辑:程序博客网 时间:2024/06/03 21:23
解题思路:
动态规划问题。
1,递归方程 steps[i] = steps[i-1] + steps[i-2]
因为从 i-1 到达 i 存在 steps(i - 1) * 1中方法, 从 i-2到达 i存在 steps(i-2) * 1 种方法,所以最后的方法数是两个相加
2,前条件:初始值,steps[0] = steps[1] = 1; 因为“站着不动” 和“走一步”都是生活的一部分
class Solution {public: int climbStairs(int n) { vector<int> steps; steps.push_back(1); steps.push_back(1); for (int i = 2; i <= n; ++i){ steps.push_back(steps[i-1] + steps[i-2]); } return steps[n]; }};
0 0
- [LeetCode]Climbing Stairs
- LeetCode: Climbing Stairs
- LeetCode: Climbing Stairs
- [LeetCode]Climbing Stairs
- LeetCode Climbing Stairs
- [Leetcode] Climbing Stairs
- Leetcode: Climbing stairs
- LeetCode Climbing Stairs
- [LeetCode] Climbing Stairs
- leetcode 107: Climbing Stairs
- [LeetCode] Climbing Stairs
- [LeetCode]Climbing Stairs
- [leetcode]Climbing Stairs
- LeetCode-Climbing Stairs
- [leetcode] Climbing Stairs
- LeetCode - Climbing Stairs
- LeetCode:Climbing Stairs
- Leetcode Climbing Stairs
- utf-8/gbk 的不可映射字符 打包后的效果:zip包 名称包含svn号
- 垂直翻页的Viewpager
- kafka(2)
- Android Studio安装及首次运行遇到的问题
- 黑马程序员----JAVA基础----IO流_2
- [LeetCode]Climbing Stairs
- 2015.07.22总结
- linux程序设计——主机字节序和网络字节序(第十五章)
- QListWidget添加小控件
- Clock类(未完善)
- pat 1023. Have Fun with Numbers (20)
- Selenium学习笔记之012:处理下拉框
- 【Linux】FrameBuffer操作入门
- HDU--2502 月之数