LeetCode 361 - Bomb Enemy

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本文介绍了一种在二维网格中使用炸弹消除敌人的算法，旨在找到放置炸弹的最佳位置，以杀死最多的敌人。算法考虑了墙壁的存在，并确保炸弹只能放置在空格上。通过四个方向的遍历，分别计算出每个位置爆炸时可以消灭的敌人数，最终找出能消灭最多敌人的位置。

题目描述：

Given a 2D grid, each cell is either a wall 'W', an enemy 'E' or empty '0' (the number zero), return the maximum enemies you can kill using one bomb.
The bomb kills all the enemies in the same row and column from the planted point until it hits the wall since the wall is too strong to be destroyed.
Note: You can only put the bomb at an empty cell.

Example:

Input: [["0","E","0","0"],["E","0","W","E"],["0","E","0","0"]]
Output: 3 
Explanation: For the given grid,

0 E 0 0 
E 0 W E 
0 E 0 0

Placing a bomb at (1,1) kills 3 enemies.

class Solution {
public:
    int maxKilledEnemies(vector<vector<char>>& grid) {
        if(grid.size()==0||grid[0].size()==0) return 0;
        int m=grid.size();
        int n=grid[0].size();
        vector<vector<int>> v1(m,vector<int>(n,0));
        vector<vector<int>> v2(m,vector<int>(n,0));
        vector<vector<int>> v3(m,vector<int>(n,0));
        vector<vector<int>> v4(m,vector<int>(n,0));
        for(int i=0;i<m;i++) { 
            for(int j=0;j<n;j++) { // v1[i][j]表示从(i,j)爆炸，向左消灭的敌人数
                if(grid[i][j]=='W') continue;
                if(grid[i][j]=='E') v1[i][j]++;
                if(j>0) v1[i][j]+=v1[i][j-1];
            }
            for(int j=n-1;j>=0;j--) { // v2[i][j]表示从(i,j)爆炸，向右消灭的敌人数
                if(grid[i][j]=='W') continue;
                if(grid[i][j]=='E') v2[i][j]++;
                if(j<n-1) v2[i][j]+=v2[i][j+1];
            }
        }
        for(int j=0;j<n;j++) { 
            for(int i=0;i<m;i++) { // v3[i][j]表示从(i,j)爆炸，向上消灭的敌人数
                if(grid[i][j]=='W') continue;
                if(grid[i][j]=='E') v3[i][j]++;
                if(i>0) v3[i][j]+=v3[i-1][j];
            }
            for(int i=m-1;i>=0;i--) { // v4[i][j]表示从(i,j)爆炸，向下消灭的敌人数
                if(grid[i][j]=='W') continue;
                if(grid[i][j]=='E') v4[i][j]++;
                if(i<m-1) v4[i][j]+=v4[i+1][j];
            }
        }
        
        int result=0;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]=='0')
                result=max(result,v1[i][j]+v2[i][j]+v3[i][j]+v4[i][j]);
            }
        }
        return result;
    }
};