62. Unique Paths

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.

Tips:机器人从左上一直走到右下,(只能走右与下)直到走到FInish的位置。

package medium;

import java.util.Arrays;

public class L62UniquePaths {
public int uniquePaths(int m, int n) {
int[][] visited = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
visited[i][j] = -1;
System.out.println(i + "," + j + ">" + visited[i][j]);
}
}
int count = movingCount(m, n, 0, 0, visited);
return count;
}
public int movingCount(int m, int n, int row, int col, int[][] visited) {
int count = 0;
if (row < 0 || col < 0 || row >= m || col >= n)
return 0;
if (row == m - 1 && col == n - 1)
return 1;
if (visited[row][col] != -1)
return visited[row][col];
count = movingCount(m, n, row + 1, col, visited) + movingCount(m, n, row, col + 1, visited);
visited[row][col] = count;
return count;
} //另外一种很快地方法。当前状态依赖于前一种状态
public int Solution2(int m, int n) {
int[] row = new int[n];
Arrays.fill(row,1);
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
row[j]+=row[j-1];
}
}
return row[n-1];
} public static void main(String[] args) {
L62UniquePaths cc = new L62UniquePaths();
int count = cc.uniquePaths(3, 4);
System.out.println(count);
}
}

 63. Unique Paths II

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
[0,0,0],
[0,1,0],
[0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

Tips:本题目是根据62题,稍作改变得来的,数组中1的位置不能走。

package medium;

public class L63UniquePaths2 {

	public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null)
return 0;
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] visited = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
visited[i][j] = -1;
System.out.println(i + "," + j + ">" + visited[i][j]);
}
}
int count = movingCount(m, n, 0, 0, visited, obstacleGrid);
return count; } public int movingCount(int m, int n, int row, int col, int[][] visited, int[][] obstacleGrid) {
int count = 0;
if (row < 0 || col < 0 || row >= m || col >= n)
return 0;
if (obstacleGrid[row][col] == 0) {
if (row == m - 1 && col == n - 1)
return 1;
if (visited[row][col] != -1)
return visited[row][col];
count = movingCount(m, n, row + 1, col, visited, obstacleGrid)
+ movingCount(m, n, row, col + 1, visited, obstacleGrid);
visited[row][col] = count;
} return count;
} public static void main(String[] args) {
L63UniquePaths2 l63 = new L63UniquePaths2();
int[][] obstacleGrid = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };
int[][] aa = { { 1 } };
int count = l63.uniquePathsWithObstacles(aa);
System.out.println(count); }
}

最新文章

  1. iOS开发之画图板(贝塞尔曲线)
  2. PIN码计算锦集
  3. HTML5 is Canvas
  4. Caffe学习系列(19): 绘制loss和accuracy曲线
  5. hdu1358 KMP
  6. 织梦DedeCMS删除所有栏目或文章后,新建ID不从1开始的解决方法
  7. TCP连接的状态详解以及故障排查
  8. 运用.NIT将数据存入数据库、读取数据库(运用封装)陈老师作业
  9. JavaScript创建命名空间、类及类成员
  10. Effective c++ 小结
  11. uva 10900
  12. 用 EasyBCD 在 Win7/8 中硬盘安装 Ubuntu
  13. React-Flux 介绍及实例演示
  14. hdu 1712 ACboy needs your help
  15. Codeforces Round #321 (Div. 2) E. Kefa and Watch 线段树hash
  16. PHP之路——MySql基础操作语句
  17. 随手写的Java向文本文件写字符串的类
  18. UGUI 帧动画插件
  19. 听翁恺老师mooc笔记(15)--文件的输入与输出
  20. python打造一个Mysql数字类型注入脚本(1)

热门文章

  1. 参考 https://raspberrypi.stackexchange.com/questions/3617/how-to-install-unrar-nonfree &gt; 1.卸载unrar-free。 $ sudo apt-get remove unrar-free \ 2.通过编辑确保您拥有源存储库/etc/apt/sources.list。 $ cat /etc/apt/sources.
  2. Centos7最小化安装之工作站设置
  3. C# 访问修饰符和const、readonly
  4. 关于第11周课堂mini dc的课堂练习
  5. c#获取已安装的所有NET版本
  6. 强化学习读书笔记 - 10 - on-policy控制的近似方法
  7. .net mvc 使用ueditor的开发(官网没有net版本?)
  8. Git生成SSH密钥
  9. SteamVR Unity Plugin - v2.0.1中的InteractionSystem
  10. 大数据-storm学习资料视频