[LeetCode62]Unique Paths
题目:
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
分类:Array ,DP
代码:
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> path(m,vector<int>(n,));
for(int i = ;i < m; ++i)
for(int j = ; j < n; ++j)
{
path[i][j] = path[i-][j] + path[i][j-];
}
return path[m-][n-];
}
};
[LeetCode62]Unique Paths的更多相关文章
- leetcode62—Unique Paths
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...
- leetcode-62. Unique Paths · DP + vector
题面 A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). ...
- Leetcode62.Unique Paths不同路径
一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为"Start" ). 机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角(在下图中标记为" ...
- [Swift]LeetCode62. 不同路径 | Unique Paths
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...
- [LeetCode] Unique Paths II 不同的路径之二
Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How m ...
- [LeetCode] Unique Paths 不同的路径
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...
- Leetcode Unique Paths II
Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How m ...
- Unique Paths II
这题在Unique Paths的基础上增加了一些obstacle的位置,应该说增加的难度不大,但是写的时候对细节的要求多了很多,比如,第一列的初始化会受到之前行的第一列的结果的制约.另外对第一行的初始 ...
- LEETCODE —— Unique Paths II [动态规划 Dynamic Programming]
唯一路径问题II Unique Paths II Follow up for "Unique Paths": Now consider if some obstacles are ...
随机推荐
- Fast portable non-blocking network programming with Libevent
Fast portable non-blocking network programming with Libevent Fast portable non-blocking network prog ...
- 我的Android学习之旅(转)
去年大概在七月份的时候误打误撞接触了一阵子Android,之后由于工作时间比较忙,无暇顾及,九月份的时候自己空闲的时间比较多,公司相对来说加班情况没以前严重.开启了个人的Android学习之旅,初衷是 ...
- java读取XML文件的四种方式
java读取XML文件的四种方式 Xml代码 <?xml version="1.0" encoding="GB2312"?> <RESULT& ...
- 碎碎念,浅饮-------Day30
这不是关于技术的文章,它偏离了我原来的计划轨迹.但,我相信这将是远远超出了技术的意义使我无论什么价格值. 高考已经开始,不知道在这片宁静的夜空下有多少人已经美美的睡了,香甜憨然.又有多少人这睡着的眼角 ...
- LAMP配置参考地址
http://www.linuxidc.com/Linux/2014-07/104563.htm
- a标签中调用js的几种方法
1. a href="javascript:js_method();" 这是我们平台上常用的方法,但是这种方法在传递this等参数的时候很容易出问题,而且javascript:协议 ...
- linux解压多个文件
方法很多种, 根据实际文件类型,位置情况进行变通: 1. for查询:for tar in *.tar.gz; do tar xvf $tar; done2. 列出文件列表,然后xargs 逐一解压: ...
- ALV DataChange EVENT
在CX项目中,根据需求,自定义一个表,维护供应商的银行账号信息,当输入供应商编号时,自动在供应商名称列里自动填写供应商名称,用到了ALV DataChange 事件 ,下面是源代码: *&- ...
- oracle nologging用法(转)
一.oracle日志模式分为(logging,force logging,nologging) 默认情况是logging,就是会记录到redo日志中,force logging是强制记录日志,nolo ...
- fragment android
在Eoe中看到了 一个关于的 详细讲解,相信对 学Fragment 有帮助 android fragment基础与源码案例: Fragment动画效果 http://www.eoeandroid.co ...