【LeetCode】Unique Paths

最新推荐文章于 2020-05-24 11:10:04 发布

原创最新推荐文章于 2020-05-24 11:10:04 发布 · 520 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 2 个专栏收录

86 篇文章

订阅专栏

动态规划

17 篇文章

订阅专栏

本文探讨了机器人从网格的左上角到达右下角的不同路径数量的问题，并通过动态规划算法提供了有效的解决方法，包括代码实现和空间优化。

参考链接

http://www.cnblogs.com/remlostime/archive/2012/11/15/2772263.html

题目描述

Unique Paths

Total Accepted: 9698 Total Submissions: 31863 My Submissions

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.

题目分析

动态规划

开一个f[m][n]的数组，f[i][j] = f[i-1][j] + f[i][j-1]，空间时间复杂度O(m*n)。用滚动数组空间复杂度可降为O(n)

代码示例

/*
编译环境CFree 5.0
博客地址：http://blog.youkuaiyun.com/Snowwolf_Yang
*/
#include 
   
    
#include 
    
     
using namespace std;
#if 0//超时 
class Solution {
public:
    int uniquePaths(int m, int n) {
    	int sum = 0;
        uniquePathsCore(m,n,0,0,sum);
        return sum;
    }
    void uniquePathsCore(int m, int n, int i, int j, int &sum)
    {
    	if(i == m-1 && j == n-1)
		{
			sum++;
			//printf("(%d,%d) ",i,j);
			//printf("end\n");
			return;
		}
    	if(j