[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 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 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
这道题让求所有不同的路径的个数,一开始还真把博主难住了,因为之前好像没有遇到过这类的问题,所以感觉好像有种无从下手的感觉。在网上找攻略之后才恍然大悟,原来这跟之前那道 Climbing Stairs 很类似,那道题是说可以每次能爬一格或两格,问到达顶部的所有不同爬法的个数。而这道题是每次可以向下走或者向右走,求到达最右下角的所有不同走法的个数。那么跟爬梯子问题一样,需要用动态规划 Dynamic Programming 来解,可以维护一个二维数组 dp,其中 dp[i][j] 表示到当前位置不同的走法的个数,然后可以得到状态转移方程为: dp[i][j] = dp[i - 1][j] + dp[i][j - 1],这里为了节省空间,使用一维数组 dp,一行一行的刷新也可以,代码如下:
解法一:
class Solution {
public:
int uniquePaths(int m, int n) {
vector<int> dp(n, );
for (int i = ; i < m; ++i) {
for (int j = ; j < n; ++j) {
dp[j] += dp[j - ];
}
}
return dp[n - ];
}
};
这道题其实还有另一种很数学的解法,参见网友 Code Ganker 的博客,实际相当于机器人总共走了 m + n - 2步,其中 m - 1 步向右走,n - 1 步向下走,那么总共不同的方法个数就相当于在步数里面 m - 1 和 n - 1 中较小的那个数的取法,实际上是一道组合数的问题,写出代码如下:
解法二:
class Solution {
public:
int uniquePaths(int m, int n) {
double num = , denom = ;
int small = m > n ? n : m;
for (int i = ; i <= small - ; ++i) {
num *= m + n - - i;
denom *= i;
}
return (int)(num / denom);
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/62
类似题目:
参考资料:
https://leetcode.com/problems/unique-paths/
https://leetcode.com/problems/unique-paths/discuss/22981/My-AC-solution-using-formula
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Unique Paths 不同的路径的更多相关文章
- [Leetcode] unique paths ii 独特路径
Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How m ...
- LeetCode: Unique Paths II 解题报告
Unique Paths II Total Accepted: 31019 Total Submissions: 110866My Submissions Question Solution Fol ...
- [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 63. Unique Paths II不同路径 II (C++/Java)
题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). ...
- [LeetCode] 62. 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每天一题】Unique Paths(唯一的路径数)
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 独特路径
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...
- LeetCode OJ:Unique Paths(唯一路径)
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 && Unique Paths II && Minimum Path Sum (动态规划之 Matrix DP )
Unique Paths https://oj.leetcode.com/problems/unique-paths/ A robot is located at the top-left corne ...
随机推荐
- Docker到底是什么?为什么它这么火!
转载来自:http://cloud.51cto.com/art/201410/453718.htm 摘要:Docker这种新的容器技术可谓热得发烫,因为有了它,人们就有可能让数量多得多的应用程序在同样 ...
- WeText项目:一个基于.NET实现的DDD、CQRS与微服务架构的演示案例
最近出于工作需要,了解了一下微服务架构(Microservice Architecture,MSA).我经过两周业余时间的努力,凭着自己对微服务架构的理解,从无到有,基于.NET打造了一个演示微服务架 ...
- C#7.0中有哪些新特性?
以下将是 C# 7.0 中所有计划的语言特性的描述.随着 Visual Studio “15” Preview 4 版本的发布,这些特性中的大部分将活跃起来.现在是时候来展示这些特性,你也告诉借此告诉 ...
- parseInt实例详解
parseInt() 函数可解析一个字符串,并返回一个整数. parseInt(string, radix) 参数 描述 string 必需.要被解析的字符串. radix 可选.表示要解析的数字的基 ...
- jquery遍历选中checkbox的id
$("[name='chkAll']:[checked]").each(function () { alert($(this).attr("id")); })
- monggodb学习系列:1,mongodb入门
http://note.youdao.com/share/?id=fa62cd2386f253af68a7e29c6638f158&type=note#/ 放在有道笔记上了,懒得复制过来,有兴 ...
- Java全角、半角字符的关系以及转换
如果搞明白了Java中全角字符和半角字符之间的关系,那他们之间的转换就不是个麻烦事儿.你只需要对这个关系有那么一个印象就足够了. 全角字符与半角字符的关系 通过下面的代码能看到Java中所有字符以及对 ...
- Java多线程--让主线程等待子线程执行完毕
使用Java多线程编程时经常遇到主线程需要等待子线程执行完成以后才能继续执行,那么接下来介绍一种简单的方式使主线程等待. java.util.concurrent.CountDownLatch 使用c ...
- Exception thrown by the agent : java.rmi.server.ExportException: Port already in use
今天有个应用一直起不来,感觉配置都对啊,奇了怪了.看日志发现如下: STATUS | wrapper | 2017/01/04 08:09:31 | Launching a JVM...INFO | ...
- 4.6 .net core依赖注入的封装
现在流行的系统一般都采用依赖注入的实现方式,利用DI容器来直接获取所用到的类/接口的实例..net core也一样采用DI的方式,提供了DI容器的接口IServiceCollection,并提供了基于 ...