博客
关于我
764. Largest Plus Sign
阅读量:427 次
发布时间:2019-03-06

本文共 4014 字,大约阅读时间需要 13 分钟。

In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An "axis-aligned plus sign of 1s of order k" has some center grid[x][y] = 1 along with 4 arms of length k-1 going up, down, left, and right, and made of 1s. This is demonstrated in the diagrams below. Note that there could be 0s or 1s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1s.

 

Examples of Axis-Aligned Plus Signs of Order k:

Order 1:000010000Order 2:0000000100011100010000000Order 3:0000000000100000010000111110000100000010000000000

 

Example 1:

Input: N = 5, mines = [[4, 2]]Output: 2Explanation:1111111111111111111111011In the above grid, the largest plus sign can only be order 2.  One of them is marked in bold.

 

Example 2:

Input: N = 2, mines = []Output: 1Explanation:There is no plus sign of order 2, but there is of order 1.

 

Example 3:

Input: N = 1, mines = [[0, 0]]Output: 0Explanation:There is no plus sign, so return 0.

 

Note:

  1. N will be an integer in the range [1, 500].
  2. mines will have length at most 5000.
  3. mines[i] will be length 2 and consist of integers in the range [0, N-1].
  4. (Additionally, programs submitted in C, C++, or C# will be judged with a slightly smaller time limit.)

 

Approach #1: Math. [Java]

class Solution {    public int orderOfLargestPlusSign(int N, int[][] mines) {        int[][] grid = new int[N][N];        for (int i = 0; i < N; ++i)            Arrays.fill(grid[i], N);                for (int[] m : mines)             grid[m[0]][m[1]] = 0;                for (int i = 0; i < N; ++i) {            for (int j = 0, k = N-1, l = 0, r = 0, u = 0, d = 0; j < N && k >= 0; ++j, --k) {                grid[i][j] = Math.min(grid[i][j], l = (grid[i][j] == 0) ? 0 : l + 1);                grid[i][k] = Math.min(grid[i][k], r = (grid[i][k] == 0) ? 0 : r + 1);                grid[j][i] = Math.min(grid[j][i], u = (grid[j][i] == 0) ? 0 : u + 1);                grid[k][i] = Math.min(grid[k][i], d = (grid[k][i] == 0) ? 0 : d + 1);            }        }                int ans = 0;                for (int i = 0; i < N; ++i) {            for (int j = 0; j < N; ++j) {                ans = Math.max(ans, grid[i][j]);            }        }                return ans;    }}

  

Analysis:

Algorithms: 

For each position (i,j) of the grid matrix, we try to extend in each of the four directions (left, right, up, down) as long as possible, then take the minimum length of 1's out of the four directions as the order of the largest axis-aligned plus sign centered at position (i, j).

 

Optimizations:

Mormally we would need a total of five matrices to make the above idea work -- one matrix for the grid itself and four more matrices for each of the four directions. However, these five matrices can be combined into one using two simple tricks:

1. For each position (i, j), we are only concerned with the minimum length of 1's out of the four directions. This implies we may combine the four matrices into one by only keeping tracking of the minimum length.

2. For each position (i, j), the order of the largest axis-aligned plus sign centered at it will be 0 if and only if grid[i][j] = 0. This implies we may further combine the grid matrix with the one obtained above.

 

Implementations:

1. Create an N-by-N matrix grid, with all elements initialized with value N.

2. Reset those elements to 0 whose positions are in the mines list.

3. For each position (i, j), find the maximum length of 1's in each of the four directions and set grid[i][j] to the minimum of these four lengths. Note that there is a simple recurrence relation the maximum length of 1's at current position with previous position for each of the four directions (labeled as l, r, u, d).

4. Loop through the grid matrix and choose the maximum element which will be the largest axis-aligned plus sign of 1's contained in the grid.

 

 

转载地址:http://futuz.baihongyu.com/

你可能感兴趣的文章
mysql 存在update不存在insert
查看>>
Mysql 学习总结(86)—— Mysql 的 JSON 数据类型正确使用姿势
查看>>
Mysql 学习总结(87)—— Mysql 执行计划(Explain)再总结
查看>>
Mysql 学习总结(88)—— Mysql 官方为什么不推荐用雪花 id 和 uuid 做 MySQL 主键
查看>>
Mysql 学习总结(89)—— Mysql 库表容量统计
查看>>
mysql 实现主从复制/主从同步
查看>>
mysql 审核_审核MySQL数据库上的登录
查看>>
mysql 导入 sql 文件时 ERROR 1046 (3D000) no database selected 错误的解决
查看>>
mysql 导入导出大文件
查看>>
mysql 将null转代为0
查看>>
mysql 常用
查看>>
MySQL 常用列类型
查看>>
mysql 常用命令
查看>>
Mysql 常见ALTER TABLE操作
查看>>
MySQL 常见的 9 种优化方法
查看>>
MySQL 常见的开放性问题
查看>>
Mysql 常见错误
查看>>
MYSQL 幻读(Phantom Problem)不可重复读
查看>>
mysql 往字段后面加字符串
查看>>
mysql 快速自增假数据, 新增假数据,mysql自增假数据
查看>>