[leetcode] 361. Bomb Enemy @ python

炸弹击杀敌军

最新推荐文章于 2019-08-02 12:53:00 发布

原创最新推荐文章于 2019-08-02 12:53:00 发布 · 638 阅读

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CC 4.0 BY-SA版权

本文介绍了一种算法问题：如何在给定的二维网格中放置一枚炸弹以杀死最多的敌人。提供了两种解决方案，一种是通过遍历每个空格计算可能击杀的敌人数量；另一种是通过预处理来减少计算复杂度。

原题

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.

解法1

遍历grid, 当遇到0时, 计算从它开始向左, 向右, 向上, 向下能杀掉的敌人个数, 并更新ans.
Time: O(mn(m+n))
Space: O(1)

代码

class Solution(object):
    def maxKilledEnemies(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        def count(x, y):
            left, right, up, down = 0, 0, 0, 0
            for i in range(x-1, -1, -1):
                if grid[i][y] == 'W':
                    break
                if grid[i][y] == 'E':
                    left += 1
                    
            for i in range(x+1, row):
                if grid[i][y] == 'W':
                    break
                if grid[i][y] == 'E':
                    right += 1
                    
            for j in range(y+1, col):
                if grid[x][j] == 'W':
                    break
                if grid[x][j] == 'E':
                    up += 1
            
            for j in range(y-1, -1, -1):
                if grid[x][j] == 'W':
                    break
                if grid[x][j] == 'E':
                    down += 1
            return left+right+up+down
        # base case
        if not grid: return 0
        row, col = len(grid), len(grid[0])
        ans = 0
        for i in range(row):
            for j in range(col):
                if grid[i][j] == '0':
                    # count the num of enemies 
                    ans = max(ans, count(i, j))
                    
        return ans

解法2

构造和grid形状相同的二维数组arr, 遍历grid4次: 向右, 向左, 向上, 向下. 每次遍历时, 使用count记录到当前位置能杀掉的敌人个数, 遇到E则递增count, 遇到W则将count设为0, 遇到0则将arr[i][j]增加count. 最后一次遍历时, 更新ans的值.

Time: O(4mn)
Space: O(1)

代码

class Solution(object):
    def maxKilledEnemies(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        # base case 
        if not grid: return 0
        row, col = len(grid), len(grid[0])
        ans = 0
        arr = [[0]*col for i in range(row)]
        # from left to right
        for i in range(row):
            count = 0
            for j in range(col):
                if grid[i][j] == 'E':
                    count += 1
                elif grid[i][j] == 'W':
                    count = 0
                else:
                    arr[i][j] += count
                    
        # from right to left
        for i in range(row):
            count = 0
            for j in range(col-1, -1, -1):
                if grid[i][j] == 'E':
                    count += 1
                elif grid[i][j] == 'W':
                    count = 0
                else:
                    arr[i][j] += count
        # from up to down
        for j in range(col):
            count = 0
            for i in range(row):
                if grid[i][j] == 'E':
                    count += 1
                elif grid[i][j] == 'W':
                    count = 0
                else:
                    arr[i][j] += count
                    
        # from down to up
        for j in range(col):
            count = 0
            for i in range(row-1, -1, -1):
                if grid[i][j] == 'E':
                    count += 1
                elif grid[i][j] == 'W':
                    count = 0
                else:
                    arr[i][j] += count
                    ans = max(ans, arr[i][j])
                    
        return ans