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.
矩阵的bfs, 考察:
1.哪些是先入队的? 外围? 还是遍历所有符合条件的? 此处改成Integer.Max_Value,
2.入队之后, 邻居元素怎么改让其入队. 怎么跳过不符合题意的元素, visited[] 是否使用
3一般用在改矩阵的题中 tips, 可以该元素e.g. 将‘o‘ 改成‘$‘(见130. Surrounded Regions)
public int[][] updateMatrix(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
Queue<int[]> queue = new LinkedList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 0) {
queue.offer(new int[]{i, j});
} else {
matrix[i][j] = Integer.MAX_VALUE;
}
}
}
int[][] dirs = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
while (!queue.isEmpty()) {
int[] cur = queue.poll();
for (int[] cell : dirs) {
int r = cur[0] + cell[0];
int c = cur[1] + cell[1];
if (r < 0 || r >= m || c < 0 || c >= n || matrix[r][c] <= matrix[cur[0]][cur[1]] + 1) {
continue;
}
matrix[r][c] = matrix[cur[0]][cur[1]] + 1;
queue.offer(new int[]{r,c});
}
}
return matrix;
}
矩阵, 常将元素构造类使用: queue.offer(new int[]{r,c}) 此处可以构造类
原文：http://www.cnblogs.com/apanda009/p/7273113.html