LWC 67: 764. Largest Plus Sign

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本文介绍了一种算法，用于寻找二维网格中最大的轴对齐加号标志。通过动态规划记录四个方向上连续1的最大数量，进而确定加号标志的大小。

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LWC 67: 764. Largest Plus Sign

传送门：764. Largest Plus Sign

Problem:

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:
000
010
000

Order 2:
00000
00100
01110
00100
00000

Order 3:
0000000
0001000
0001000
0111110
0001000
0001000
0000000

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.)

思路：
动态规划，分别记录4个方向上的最大连续1的个数。比如”1001111”，每个位置出现的最大连续1的个数分别为：”1001234”，有了4个方向的最长连续1，order就是这四个方向的最小值，遍历每个位置的order，求出最大order即可。

Java版本：

    public int orderOfLargestPlusSign(int N, int[][] mines) {
        int[][][] grid = new int[N][N][4];
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                for (int k = 0; k < 4; ++k) {
                    grid[i][j][k] = 1;
                }
            }
        }

        for (int[] mine : mines) {
            int r = mine[0];
            int c = mine[1];
            for (int k = 0; k < 4; ++k) grid[r][c][k] = 0;
        }

        for (int i = 0; i < N; ++i) {
            for (int j = 1; j < N; ++j) {
                if (grid[i][j][0] == 1)
                    grid[i][j][0] += grid[i][j - 1][0]; 
            }
            for (int j = N - 2; j >= 0; --j) {
                if (grid[i][j][1] == 1)
                    grid[i][j][1] += grid[i][j + 1][1];
            }
        }

        for (int j = 0; j < N; ++j) {
            for (int i = 1; i < N; ++i) {
                if (grid[i][j][2] == 1)
                    grid[i][j][2] += grid[i - 1][j][2];
            }
            for (int i = N - 2; i >= 0; --i) {
                if (grid[i][j][3] == 1)
                    grid[i][j][3] += grid[i + 1][j][3];
            }
        }

        int ans = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                int order = Math.min(Math.min(grid[i][j][0], grid[i][j][1]), Math.min(grid[i][j][2], grid[i][j][3]));
                ans = Math.max(ans, order);
            }
        }
        return ans;
    }

Python版本：

class Solution(object):
    def orderOfLargestPlusSign(self, N, mines):
        """
        :type N: int
        :type mines: List[List[int]]
        :rtype: int
        """
        grid = [[1] * N for _ in range(N)]
        lf = [[0] * N for _ in range(N)]
        dn = [[0] * N for _ in range(N)]
        rt = [[0] * N for _ in range(N)]
        up = [[0] * N for _ in range(N)]

        for i, j in mines: grid[i][j] = 0

        for i in xrange(N):
            for j in xrange(N):
                if grid[i][j] == 1:
                    lf[i][j] = 1 if j == 0 else lf[i][j - 1] + 1
                if grid[j][i] == 1:
                    dn[j][i] = 1 if j == 0 else dn[j - 1][i] + 1
        for i in xrange(N):
            for j in xrange(N - 1, -1, -1):
                if grid[i][j] == 1:
                    rt[i][j] = 1 if j == N - 1 else rt[i][j + 1] + 1
                if grid[j][i] == 1:
                    up[j][i] = 1 if j == N - 1 else up[j + 1][i] + 1

        ans = 0
        for i in xrange(N):
            for j in xrange(N):
                order = min(lf[i][j], rt[i][j], up[i][j], dn[i][j])
                ans = max(ans, order)
        return ans