标签：题目 list space solution cee nbsp NPU put get
原题链接在这里：https://leetcode.com/problems/01-matrix/description/
题目：
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:
题解：
先把非0点标记成Integer.MAX_VALUE. 然后从每个0点开始做BFS. 若是对周围的点更新出更小的值就把更小值放入queue中用作将来的BFS.
Time Complexity: O(m*n). m = matrix.length, n = matrix[0].length. 每个点最多进入queue 4 次.
Space: O(m*n). queue size.
AC Java:
1 class Solution { 2 public int[][] updateMatrix(int[][] matrix) { 3 if(matrix == null || matrix.length == 0 || matrix[0].length == 0){ 4 return matrix; 5 } 6 7 int m = matrix.length; 8 int n = matrix[0].length; 9 10 LinkedList<int []> que = new LinkedList<int []>(); 11 for(int i = 0; i<m; i++){ 12 for(int j = 0; j<n; j++){ 13 if(matrix[i][j] == 0){ 14 que.add(new int[]{i, j}); 15 }else{ 16 matrix[i][j] = Integer.MAX_VALUE; 17 } 18 } 19 } 20 21 int [][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; 22 23 while(!que.isEmpty()){ 24 int [] cur = que.poll(); 25 for(int [] dir : dirs){ 26 int dx = cur[0] + dir[0]; 27 int dy = cur[1] + dir[1]; 28 if(dx<0 || dx>=m || dy<0 || dy>=n || matrix[cur[0]][cur[1]]+1>=matrix[dx][dy]){ 29 continue; 30 } 31 32 matrix[dx][dy] = matrix[cur[0]][cur[1]]+1; 33 que.add(new int[]{dx, dy}); 34 } 35 } 36 37 return matrix; 38 } 39 }
标签：题目 list space solution cee nbsp NPU put get
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/9602413.html
