553 · 炸弹袭击

描述

给定一个二维矩阵, 每一个格子可能是一堵墙 W,或者 一个敌人 E 或者空 0 (数字 '0'), 返回你可以用一个炸弹杀死的最大敌人数. 炸弹会杀死所有在同一行和同一列没有墙阻隔的敌人。 由于墙比较坚固，所以墙不会被摧毁.

你只能在空的地方放置炸弹.

样例

样例1

输入:
grid =[
     "0E00",
     "E0WE",
     "0E00"
]
输出: 3
解释:
把炸弹放在 (1,1) 能杀3个敌人

样例2

输入:
grid =[
     "0E00",
     "EEWE",
     "0E00"
]
输出: 2
解释:
P把炸弹放在 (0,0) 或 (0,3) 或 (2,0) 或 (2,3) 能杀2个敌人

思路：通过4种方向的动态规划，解决问题 

递推公式：

        1.当前位置是wall，dp[i][j] = 0;

        2.当前位置是enemy，dp[i][j]=前一种状态+1;

        3.当前位置是0，dp[i][j] = 前一种状态;

public class Solution {
    /**
     * @param grid: Given a 2D grid, each cell is either 'W', 'E' or '0'
     * @return: an integer, the maximum enemies you can kill using one bomb
     */
    public int maxKilledEnemies(char[][] grid) {
        // write your code here
        if(grid.length==0) return 0;
        int m = grid.length;
        int n = grid[0].length;
        int[][] up = new int[m][n];
        int[][] down = new int[m][n];
        int[][] left = new int[m][n];
        int[][] right = new int[m][n];
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                int a = m-i-1;
                int b = n-j-1;
                if(i==0)
                {
                    if(grid[i][j]=='E') up[i][j] = 1;
                    else up[i][j] = 0;
                }
                else{
                    if(grid[i][j]=='E') up[i][j] = up[i-1][j]+1;
                    else if(grid[i][j]=='W') up[i][j] =0;
                    else up[i][j] = up[i-1][j]; 
                }

                if(i==0)
                {
                    if(grid[a][j]=='E') down[a][j] = 1;
                    else down[a][j] = 0;
                }
                else{
                    if(grid[a][j]=='E') down[a][j] = down[a+1][j]+1;
                    else if(grid[a][j]=='W') down[a][j] =0;
                    else down[a][j] = down[a+1][j]; 
                }

                if(j==0)
                {
                    if(grid[i][j]=='E') left[i][j] = 1;
                    else left[i][j] = 0;
                }
                else{
                    if(grid[i][j]=='E') left[i][j] = left[i][j-1]+1;
                    else if(grid[i][j]=='W') left[i][j] =0;
                    else left[i][j] = left[i][j-1]; 
                }

                if(j==0)
                {
                    if(grid[i][b]=='E') right[i][b] = 1;
                    else right[i][b] = 0;
                }
                else{
                    if(grid[i][b]=='E') right[i][b] = right[i][b+1]+1;
                    else if(grid[i][b]=='W') right[i][b] =0;
                    else right[i][b] = right[i][b+1]; 
                }
            

            }
        }
        int x=-1,y=-1;
        int[][] arr = new int[m][n];
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                arr[i][j] = up[i][j] +down[i][j] +right[i][j]+left[i][j]; 
                //System.out.println(up[i][j] +" "+down[i][j]+" "+right[i][j]+" "+left[i][j]);
            }
        }

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='0'){
                    if(x==-1){
                        x=i;
                        y=j;
                    }else if(arr[x][y]<arr[i][j]){
                        x = i;
                        y=j;
                    }
                }
                
            }
        }
        if(x==-1) return 0;
        return arr[x][y];
    }
}

标签：int,553,length,炸弹,grid,0E00,new,袭击
来源： https://blog.csdn.net/weixin_45845865/article/details/121545735