解题思路
「多源 BFS 问题」先统计所有新鲜橘子数量，同时将初始所有腐烂橘子作为BFS第一层起点，然后按每分钟扩散感染相邻新鲜橘子，每感染一个就减少新鲜计数，直到没有新鲜橘子或无法继续扩散，最终返回耗时或 -1
参考代码
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| class Solution {
private static final int[][] DIRECTIONS = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public int orangesRotting(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int fresh = 0;
List<int[]> q = new ArrayList<>();
for(int i = 0; i < m; i ++) {
for(int j = 0; j < n; j ++) {
if(grid[i][j] == 1) {
fresh ++;
} else if(grid[i][j] == 2) {
q.add(new int[]{i, j});
}
}
}
int res = 0;
while(fresh > 0 && !q.isEmpty()) {
res ++;
List<int[]> tmp = q;
q = new ArrayList<>();
for(int[] pos : tmp) {
for(int[] d: DIRECTIONS) {
int i = pos[0] + d[0];
int j = pos[1] + d[1];
if(0 <= i && i < m && 0 <= j && j < n && grid[i][j] == 1) {
fresh --;
grid[i][j] = 2;
q.add(new int[]{i, j});
}
}
}
}
return fresh > 0 ? -1 : res;
}
}
|