[Leetcode] 764. Largest Plus Sign 解题报告
题目：
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.)
思路：
1) 我们定义一个N*N的矩阵grid，将其每一个元素初始化为N，并且将mines列表中的元素置为0；
2）对于每个位置(i, j)，我们分别在四个方向上找出最大值，然后将grid[i][j]的值置为四个方向上的最大值的最小者。我们需要注意的是，在每个方向上的最大值其实和它周围的元素的最大值是有递推关系的，我们利用这些递推关系可以实现时间复杂度的降低。
3）通过循环，找出grid中的最大值，即为题目所求。
该算法的时间复杂度为O(n^2)，空间复杂度也是O(n^2)。
代码：
class Solution {
public:
int orderOfLargestPlusSign(int N, vector>& mines) {
vector> grid(N, vector(N, N));
for (auto& m : mines) { // initialize the elements in grid
grid[m[0]][m[1]] = 0;
}
for (int i = 0; i < N; ++i) {
int l = 0, r = 0, u = 0, d = 0;
for (int j = 0, k = N - 1; j < N; j++, k--) {
grid[i][j] = min(grid[i][j], l = (grid[i][j] == 0 ? 0 : l + 1));
grid[i][k] = min(grid[i][k], r = (grid[i][k] == 0 ? 0 : r + 1));
grid[j][i] = min(grid[j][i], u = (grid[j][i] == 0 ? 0 : u + 1));
grid[k][i] = min(grid[k][i], d = (grid[k][i] == 0 ? 0 : d + 1));
}
}
int res = 0; // find the largest element in grid
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
res = max(res, grid[i][j]);
}
}
return res;
}
};