Leetcode 361. Bomb Enemy (Medium) (cpp)

最新推荐文章于 2019-09-20 17:07:29 发布

Niko_Ke

最新推荐文章于 2019-09-20 17:07:29 发布

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本文提供了一种使用动态规划解决LeetCode第361题的方法，该题目标是在一个二维网格中放置一枚炸弹以消灭最多的敌人。文章详细介绍了如何通过预先计算上下左右方向上的敌人数量来高效解决问题。

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Leetcode 361. Bomb Enemy (Medium) (cpp)

Tag: Dynamic Programming

Difficulty: Medium

/*

361. Bomb Enemy (Medium)

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 that you can only put the bomb at an empty cell.

Example:
For the given grid

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

return 3. (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 col = grid[0].size(), row = grid.size();
		vector<vector<int>> vrl(row, vector<int>(col)), vbu(row, vector<int>(col));
		for (int j = col - 1; j >= 0; j--) {
			for (int i = row - 1; i >= 0; i--) {
				if (grid[i][j] != 'W') {
					int e = grid[i][j] == 'E';
					vrl[i][j] = j == col - 1 ? e : e + vrl[i][j + 1];
					vbu[i][j] = i == row - 1 ? e : e + vbu[i + 1][j];
				}
			}
		}
		vector<int> vlr(row), vub(col);
		int res = 0;
		for (int i = 0; i < row; i++) {
			for (int j = 0; j < col; j++) {
				if (grid[i][j] == '0') {
					res = max(vlr[i] + vrl[i][j] + vub[j] + vbu[i][j], res);
				}
				else {
					vub[j] = grid[i][j] == 'W' ? 0 : vub[j] + 1;
					vlr[i] = grid[i][j] == 'W' ? 0 : vlr[i] + 1;
				}
			}
		}
		return res;
	}
};