LeetCode Unique Paths
来源:互联网 发布:中咨数据有限公司官网 编辑:程序博客网 时间:2024/06/05 22:49
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.
思路分析:这题开始想到用DFS 搜索来做,一般搜索从S到F的路径一边count路径的数量,但是算法时间复杂度太高,用DFS算法时间复杂度为O(2^(m+n)),是指数级复杂度。其实,这题可以用DP得到更快的算法,具体而言,定义m*n维二维数组dp[][],dp[i][j]表示从s出发到达格子(i,j)的路径数目,现在我们可以思考如果写dp方程。通过观察网格特征和走法,我们可以分析得知,到达(i,j)只有两种走法,第一是从(i-1,j)向下到(i,j),第二是从(i,j-1)向右到(i,j),而dp[i][j-1]和dp[i-1][j]是已经保存的中间计算结果,所以可以得到dp递推方程为dp[i][j] = dp[i][j-1] + dp[i-1][j]。所以可以从(0,0)开始不断计算到右侧和下方的格子的路径总数,直到算出dp[m-1][n-1],算法时间复杂度为O(mn),空间复杂度也为O(mn)。其实还可以进一步节省空间,因为计算dp[i][j]时只需要知道相邻左侧一列或者上面一行的计算结果,因此我们只需要定义一个数组row或者col(取决于m和n哪个更小),然后也可以不断推算出dp[m-1][n-1].
AC Code
public class Solution { public int uniquePaths(int m, int n) { //dp[i][j] = dp[i-1][j] + dp[i][j-1]; int [][] dp = new int[m][n]; for(int i = 0; i < m; i++){ dp[i][0] = 1; } for(int j = 0; j < n; j++){ dp[0][j] = 1; } for(int i = 1; i < m; i++){ for(int j = 1; j< n; j++){ dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; }}
- leetcode Unique Paths && Unique Paths
- Leetcode-Unique Paths&Unique PathsII
- [LeetCode] Unique Paths and Unique Paths II
- leetcode--Unique Paths && Unique Paths ii
- leetcode--Unique Paths && Unique Paths ii
- 【Leetcode】【python】Unique Paths/Unique Paths II
- LeetCode: Unique Paths
- LeetCode: Unique Paths II
- [Leetcode] Unique Paths
- [Leetcode] Unique Paths II
- LeetCode : Unique Paths
- LeetCode : Unique Paths II
- leetcode 9: Unique Paths
- [LeetCode] Unique Paths
- [LeetCode] Unique Paths II
- [LeetCode]Unique Paths
- [LeetCode]Unique Paths II
- [LeetCode] Unique Paths
- ios开发-设置view背景
- 纯CSS实现3D图像轮转
- 数据库存储过程
- 纯CSS仿制Google女生节Doodle
- 数据库-触发器
- LeetCode Unique Paths
- webview去掉点击按钮的边框
- Android模拟器使用的快捷键、参数、注意事项
- JDBC中的预编译
- String、Stringbuilder、StringBuffer的区别
- IIS中MIME类型映射汇总
- tomcat忘了用户名和密码怎么办
- 数据挖掘算法
- 职业教育利益相关者的相关论述及研究不足