题目描述：
LeetCode 764. Largest Plus Sign
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:
Nwill be an integer in the range[1, 500].mineswill have length at most5000.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.)
题目大意：
二维方阵grid长宽为N，初始为全0矩阵。给定位置数组mines，在grid中将mines中的各位置设为1。
求grid中“十字形全1区域”的最大长度，关于“十字形全1区域”的定义详见测试用例。
解题思路：
时间复杂度O(N^2)
用O(N^2)的代价求出每一行的“一字型全1区域”的长度 用O(N^2)的代价求出每一列的“一字型全1区域”的长度 遍历取最小值即为“十字形全1区域”的长度
Java代码：
class Solution {
public int orderOfLargestPlusSign(int N, int[][] mines) {
int[][] cnts = new int[N][N];
int[][] grid = new int[N][N];
for (int[] mine : mines) {
grid[mine[0]][mine[1]] = 1;
}
for (int i = 0; i < N; i++) {
int[] cntLR = new int[N + 1];
for (int j = 0; j < N; j++) {
cntLR[j] = grid[i][j] == 0 ? j > 0 ? cntLR[j - 1] + 1 : 1 : 0;
}
int cntRL = 0;
for (int j = N - 1; j > -1; j--) {
cntRL = grid[i][j] == 0 ? cntRL + 1 : 0;
cnts[i][j] = Math.min(cntLR[j], cntRL);
}
}
int ans = 0;
for (int i = 0; i < N; i++) {
int[] cntUD = new int[N + 1];
for (int j = 0; j < N; j++) {
cntUD[j] = grid[j][i] == 0 ? j > 0 ? cntUD[j - 1] + 1 : 1 : 0;
}
int cntDU = 0;
for (int j = N - 1; j > -1; j--) {
cntDU = grid[j][i] == 0 ? cntDU + 1 : 0;
cnts[j][i] = Math.min(cnts[j][i], Math.min(cntUD[j], cntDU));
ans = Math.max(ans, cnts[j][i]);
}
}
return ans;
}
}
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