[Swift]LeetCode764. 最大加号标志 | Largest Plus Sign

最新推荐文章于 2021-12-07 11:28:21 发布

转载最新推荐文章于 2021-12-07 11:28:21 发布 · 183 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/strengthen/p/10533262.html

文章标签：

#swift #c# #runtime

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10533262.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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:

Example 1:

Input: N = 5, mines = [[4, 2]]
Output: 2
Explanation:
11111
11111
11111
11111
11011
In 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: 1
Explanation:
There is no plus sign of order 2, but there is of order 1.

Example 3:

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

Note:

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

在一个大小在 (0, 0) 到 (N-1, N-1) 的2D网格 grid 中，除了在 mines 中给出的单元为 0，其他每个单元都是 1。网格中包含 1 的最大的轴对齐加号标志是多少阶？返回加号标志的阶数。如果未找到加号标志，则返回 0。

一个 k" 阶由 1 组成的“轴对称”加号标志具有中心网格 grid[x][y] = 1 ，以及4个从中心向上、向下、向左、向右延伸，长度为 k-1，由 1 组成的臂。下面给出 k" 阶“轴对称”加号标志的示例。注意，只有加号标志的所有网格要求为 1，别的网格可能为 0 也可能为 1。

k 阶轴对称加号标志示例:

示例 1：

输入: N = 5, mines = [[4, 2]]
输出: 2
解释:
11111
11111
11111
11111
11011
在上面的网格中，最大加号标志的阶只能是2。一个标志已在图中标出。

示例 2：

输入: N = 2, mines = []
输出: 1
解释:
11
11
没有 2 阶加号标志，有 1 阶加号标志。

示例 3：

输入: N = 1, mines = [[0, 0]]
输出: 0
解释:
0
没有加号标志，返回 0 。

提示：

整数N 的范围： [1, 500].
mines 的最大长度为 5000.
mines[i] 是长度为2的由2个 [0, N-1] 中的数组成.
(另外,使用 C, C++, 或者 C# 编程将以稍小的时间限制进行判断.)

Runtime: 1368 ms

Memory Usage: 19.8 MB

 1 class Solution {
 2     func orderOfLargestPlusSign(_ N: Int, _ mines: [[Int]]) -> Int {
 3         var res:Int = 0
 4         var dp:[[Int]] = [[Int]](repeating:[Int](repeating:N,count:N),count:N)
 5         for mine in mines
 6         {
 7             dp[mine[0]][mine[1]] = 0
 8         }
 9         for i in 0..<N
10         {
11             var l:Int = 0
12             var r:Int = 0
13             var u:Int = 0
14             var d:Int = 0
15             var j:Int = 0
16             var k:Int = N - 1
17             while(j < N)
18             {
19                 l = dp[i][j] != 0 ? l + 1 : 0
20                 dp[i][j] = min(dp[i][j],l)
21                 u = dp[j][i] != 0 ? u + 1 : 0
22                 dp[j][i] = min(dp[j][i],u)
23                 r = dp[i][k] != 0 ? r + 1 : 0
24                 dp[i][k] = min(dp[i][k],r)
25                 d = dp[k][i] != 0 ? d + 1 : 0
26                 dp[k][i] = min(dp[k][i],d)
27                                
28                 j += 1
29                 k -= 1
30             }
31         }
32         for k in 0..<(N * N)
33         {
34             res = max(res, dp[k / N][k % N])
35         }
36         return res
37     }
38 }