深信服9.23日Java笔试
深信服
分享一道「新冠防控」，最后没时间了，思路大致是判断每个元素周围的四个方向上是否存在1，存在的话置当前置为2，count++就完事了
import java.util.*;
public class Solution4 {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param grid int整型二维数组
* @return int整型
*/
static int count = 0;
public static void main(String[] args) {
int[][] arr = {{0,1,0,0,1},{0,1,0,0,1},{0,0,0,0,1},{0,0,0,0,0},{0,0,0,0,0},};
Solution4 s = new Solution4();
int n = s.ncov_defect(arr);
System.out.println(count);
}
public int ncov_defect (int[][] grid) {
// write code here
int row = grid.length;
int col = grid[0].length;
int count = 0;
for (int i = 0; i < row; i++){
for (int j = 0; j < col; j++){
dfs(grid, i, j); // 修改2的个数
}
}
return count;
}
public void dfs(int[][] grid, int row, int col){
if(!inArea(grid, row, col)){
return;
}
if (grid[row][col] == 1 || grid[row][col] == 2){
// 如果等于1则返回
return;
}
if (inArea(grid,row, col - 1) && grid[row][col- 1] == 1 ){
grid[row][col] = 2;
count++;
}
if ( inArea(grid,row, col + 1) && grid[row][col+1] == 1 ){
grid[row][col] = 2;
count++;
}
if (inArea(grid,row -1, col ) && grid[row - 1][col] == 1 ){
grid[row][col] = 2;
count++;
}
if ( inArea(grid,row +1, col )&& grid[row + 1][col] == 1 ){
grid[row][col] = 2;
count++;
}
}
public boolean inArea(int[][] grid,int row, int col){
if(0 <= row && row < grid.length && 0 <= col && col < grid[0].length){
return true;
}else{
return false;
}
}
}
