LWC 67: 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：
代码语言：javascript复制 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：
代码语言：javascript复制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
