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 1
s contained in the grid? Return the order of the plus sign. If there is none, return 0.
An "axis-aligned plus sign of 1
s 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 1
s. This is demonstrated in the diagrams below. Note that there could be 0
s or 1
s 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 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.)
A:
首先上来,想每个点,继续走更深一步。如果不能走,则把该点标记为废物点(负值)
class Solution { public: int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { vector<vector<int>> V(N, vector<int>(N, 1)); for (auto pos : mines) { V[pos[0]][pos[1]] = 0; } int radius = 0; // next possible radius. bool tryMore = true; while (tryMore) { tryMore = false; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (V[i][j] > 0) { if (i - radius >= 0 && V[i-radius][j] != 0 && i + radius < N && V[i+radius][j] != 0 && j - radius >= 0 && V[i][j-radius] != 0 && j + radius < N && V[i][j+radius] != 0) { tryMore = true; V[i][j] = radius + 1; }else { V[i][j] = -1; } } } } if (tryMore) radius++; } return radius; // should be radius + 1(for pixel itself) - 1(for next possible) } };
哎,, 还是 LTE了
继续用DP 方法
既然从1 开始走,不行,那么就考虑从Input给的是mines 是0 的位置考虑
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