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 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)
解法一：建立四个累加数组v1, v2, v3, v4，其中v1是水平方向从左到右的累加数组，v2是水平方向从右到左的累加数组，v3是竖直方向从上到下的累加数组，v4是竖直方向从下到上的累加数组，我们建立好这个累加数组后，对于任意位置(i, j)，其可以炸死的最多敌人数就是v1[i][j] + v2[i][j] + v3[i][j] + v4[i][j]，最后我们通过比较每个位置的累加和，就可以得到结果
class Solution():
def maxKilledEnemies(self, grid):
# write your code here
if not grid:return 0
m, n = len(grid), len(grid[0])
v1 = [[0 for _ in grid[0]] for _ in grid]
v2 = [[0 for _ in grid[0]] for _ in grid]
v3 = [[0 for _ in grid[0]] for _ in grid]
v4 = [[0 for _ in grid[0]] for _ in grid]
res = 0
#print(m,n)
for i in range(m):
for j in range(n):
t = 0 if j == 0 or grid[i][j] == 'W' else v1[i][j-1]
v1[i][j] = t+1 if grid[i][j] == 'E' else t
for j in range(n)[::-1]:
t = 0 if j == n-1 or grid[i][j] == 'W' else v2[i][j+1]
v2[i][j] = t+1 if grid[i][j] == 'E' else t
for j in range(n):
for i in range(m):
t = 0 if i == 0 or grid[i][j] == 'W' else v3[i-1][j]
v3[i][j] = t+1 if grid[i][j] == 'E' else t
for i in range(m)[::-1]:
t = 0 if i == m-1 or grid[i][j] == 'W' else v4[i+1][j]
v4[i][j] = t+1 if grid[i][j] == 'E' else t
print(v3)
for i in range(m):
for j in range(n):
if grid[i][j] == '0':
res = max(res, v1[i][j]+v2[i][j]+v3[i][j]+v4[i][j])
return res
a = Solution()
L = ["0E00","E0WE","0E00"]
print(a.maxKilledEnemies(L))解法二：这种解法比较省空间，写法也比较简洁，需要一个rowCnt变量，用来记录到下一个墙之前的敌人个数。还需要一个数组colCnt，其中colCnt[j]表示第j列到下一个墙之前的敌人个数。算法思路是遍历整个数组grid，对于一个位置grid[i][j]，对于水平方向，如果当前位置是开头一个或者前面一个是墙壁，我们开始从当前位置往后遍历，遍历到末尾或者墙的位置停止，计算敌人个数。对于竖直方向也是同样，如果当前位置是开头一个或者上面一个是墙壁，我们开始从当前位置向下遍历，遍历到末尾或者墙的位置停止，计算敌人个数。可能会有人有疑问，为啥rowCnt就可以用一个变量，而colCnt就需要用一个数组呢，为啥colCnt不能也用一个变量呢？原因是由我们的遍历顺序决定的，我们是逐行遍历的，在每行的开头就统计了该行的敌人总数，所以在该行遍历没必要用数组，但是每次移动时就会换到不同的列，我们总不能没每个列就重新统计一遍吧，所以就在第一行时一起统计了存到数组中供后来使用。有了水平方向和竖直方向敌人的个数，那么如果当前位置是0，表示可以放炸弹，我们更新结果res即可，参见代码如下：
def maxKilledEnemies(self, grid):
if not grid:return 0
m, n = len(grid), len(grid[0])
res, rowCnt= 0, 0
colCnt = [0 for _ in range(n)]
for i in range(m):
for j in range(n):
if j == 0 or grid[i][j-1] == 'W':
rowCnt = 0
for k in range(j, n):
if grid[i][k] != 'W':
rowCnt += (grid[i][k] == 'E')
else:
break
if i == 0 or grid[i-1][j] == 'W':
colCnt[j] = 0
for k in range(i, m):
if grid[k][j] != 'W':
colCnt[j] += (grid[k][j] == 'E')
else:
break
if grid[i][j] == '0':
res = max(res, rowCnt+colCnt[j])
return res