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_542.java
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_542.java
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package com.fishercoder.solutions;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;
/**
* 542. 01 Matrix
*
* Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
The distance between two adjacent cells is 1.
Example 1:
Input:
0 0 0
0 1 0
0 0 0
Output:
0 0 0
0 1 0
0 0 0
Example 2:
Input:
0 0 0
0 1 0
1 1 1
Output:
0 0 0
0 1 0
1 2 1
Note:
The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
The cells are adjacent in only four directions: up, down, left and right.
*/
public class _542 {
public static class Solution1 {
public List<List<Integer>> updateMatrix(List<List<Integer>> matrix) {
int m = matrix.size();
int n = matrix.get(0).size();
Deque<int[]> deque = new LinkedList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix.get(i).get(j) == 0) {
deque.offer(new int[]{i, j});
} else {
matrix.get(i).set(j, Integer.MAX_VALUE);
}
}
}
final int[] dirs = new int[]{0, 1, 0, -1, 0};
while (!deque.isEmpty()) {
int[] currentCell = deque.poll();
for (int i = 0; i < dirs.length - 1; i++) {
int nextRow = currentCell[0] + dirs[i];
int nextCol = currentCell[1] + dirs[i + 1];
if (nextRow < 0 || nextCol < 0 || nextRow >= m || nextCol >= n || matrix.get(nextRow).get(nextCol) <= matrix.get(currentCell[0]).get(currentCell[1]) + 1) {
continue;
}
deque.offer(new int[]{nextRow, nextCol});
matrix.get(nextRow).set(nextCol, matrix.get(currentCell[0]).get(currentCell[1]) + 1);
}
}
return matrix;
}
}
}