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

题目大意:

从m乘n矩阵左上角走到右下角,每一步只能向右或向下,求共有多少条不同路径。

递归解决:

 class Solution {
public:
vector<vector<int>> v; int solve(int m, int n) {//从(m, n)到(1, 1)有多少条路径
if (m == || n == )
return ;
if (v[m][n] >= )
return v[m][n];
v[m][n] = solve(m - , n) + solve(m, n - );
return v[m][n];
} int uniquePaths(int m, int n) {
if (m == || n == )
return ;
if (m == || n == )
return ;
v.resize(m + , vector<int>(n + , -));
return solve(m - , n) + solve(m, n - );
}
};

迭代:

 class Solution {
public: int uniquePaths(int m, int n) {
if (m == || n == )
return ;
vector<vector<int>> v(m, vector<int>(n));
int i, j;
for (i = ; i < m; i++) {
for (j = ; j < n; j++) {
if (i == || j == )
v[i][j] = ;
else
v[i][j] = v[i - ][j] + v[i][j - ];
}
}
return v[m - ][n - ];
}
};

leetcode 62、Unique Paths的更多相关文章

  1. leetcode@ [62/63] Unique Paths II

    class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleG ...

  2. &lt;LeetCode OJ&gt; 62. / 63. Unique Paths(I / II)

    62. Unique Paths My Submissions Question Total Accepted: 75227 Total Submissions: 214539 Difficulty: ...

  3. 【leetcode】62.63 Unique Paths

    62. Unique Paths A robot is located at the top-left corner of a m x n grid (marked 'Start' in the di ...

  4. LeetCode(62)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). ...

  5. 【一天一道LeetCode】#63. Unique Paths II

    一天一道LeetCode (一)题目 Follow up for "Unique Paths": Now consider if some obstacles are added ...

  6. 【LeetCode】63. Unique Paths II 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/unique-pa ...

  7. 【LeetCode练习题】Unique Paths II

    Unique Paths II Follow up for "Unique Paths": Now consider if some obstacles are added to ...

  8. 【LeetCode练习题】Unique Paths

    Unique Paths A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagra ...

  9. LeetCode OJ 63. Unique Paths II

    Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How m ...

随机推荐

  1. [转] Jenkins Pipeline插件十大最佳实践

    [From] http://blog.didispace.com/jenkins-pipeline-top-10-action/ Jenkins Pipeline 插件对于 Jenkins 用户来说可 ...

  2. Spring中如何向 Bean注入系统属性或环境变量

    [转自] http://unmi.cc/spring-injection-system-properties-env/ 在 Spring 中为 javabean 注入属性文件中的属性值一般人都知道的, ...

  3. poj3187

    一.题意:给定n,求1~n的一个排列,这个排列需要满足以下两个要求:1.杨辉三角最后的和为sum  2.字典序最小 二.思路:暴力枚举每一个排列,然后计算和并与sum进行比较.这里我比较费解的是为什么 ...

  4. java中的线程(3):线程池类 ThreadPoolExecutor「线程池的类型、参数、扩展等」

    官方文档: https://docs.oracle.com/javase/7/docs/api/java/util/concurrent/ThreadPoolExecutor.html 1.简介 pu ...

  5. 2.4 GO Interface

    package itface type Sender interface { Send(url string) string } type Geter interface { Get(url stri ...

  6. 页面跳转问题-button 确定提交按钮

    form和ajax不可一起用了,button标签默认是用的form表单,所以导致跳转有问题,form不能和ajax一起用的,切记

  7. Unity 判断Animatior是否播放完

    public Animator animator; void Start() { animator = this.GetComponent<Animator>(); } void Upda ...

  8. zookeeper 编程框架 curator

    Curator框架提供了一套高级的API, 简化了ZooKeeper的操作. 它增加了很多使用ZooKeeper开发的特性,可以处理ZooKeeper集群复杂的连接管理和重试机制. 这些特性包括: 自 ...

  9. import java.util.Collections类

    Collections类提供了一些操作集合的方法  下面介绍几个方法 1.将集合变为线程安全的 三个方法分别对应了ArrayList,HashMap,HashSet: Collections.sync ...

  10. 从零实现一个简易jQuery框架之一—jQuery框架概述

    我们知道,不管学习任何一门框架,了解其设计的理念.目的.总体的结构及核心特性对我们使用和后续的深入理解框架都是有很大的帮助的.因此在这里先梳理一下本人对jQuery框架的一些理解. 设计目的(为什么要 ...