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)
Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.
方法一：
这道题的思想就是遍历数组中的每一个点，当前点能够得到的最大值为从左边的头（边界或wall）到右边的头（遇到边界或wall）的值（row值）＋从上面的头（边界或wall）到下面的头（遇到边界或wall）的值（col值）。每一个row范围内的点都共用当前的row值，每一个col范围内的点都共用当前col值。当从新的边界或者wall开始时，相当于进入了新的一段范围，要重新计算row值或者col值。
1.遍历数组中每一个点，若为0则开始计算
2.若当前点为第一列或者左边一个点为wall，表明进入了一个新的区间，需要重新计算。从该点开始一直向右直到遇到边界或者wall，在该过程中，每遇到一个E就将row值＋1+
3.若当前点为第一行或者上边一个点为wall，表明进入了一个新的区间，需要重新计算。从该点开始一直向下直到遇到边界或者wall，在该过程中，每遇到一个E就将col值＋1
4.重复2-3步骤
方法二：
上面的方法，会存在大量的重复计算。好的解法是用DP，用一个cache来存下部分结果来避免重复运算。这里用一个m*2的数组enemyCount来储存每一行的结果。enemyCount[i][0]储存了第i行某一个W的位置，enemyCount[i][1]储存了enemyCount[i][0]所标识的W到之前一个W之间E的个数，初始化：enemyCount[i][0] = -1，enemyCount[i][1] = 0。
Java: Iteration
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) {
if(grid == null || grid.length == 0 || grid[0].length == 0){
return 0;
}
int m = grid.length;
int n = grid[0].length;
int result = 0;
int rows = 0;
int[] cols = new int[n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (j == 0 || grid[i][j-1] == ‘W‘) {
rows = 0;
for (int k = j; k < n && grid[i][k] != ‘W‘; ++k)
if (grid[i][k] == ‘E‘)
rows += 1;
}
if (i == 0 || grid[i-1][j] == ‘W‘) {
cols[j] = 0;
for (int k = i; k < m && grid[k][j] != ‘W‘; ++k)
if (grid[k][j] == ‘E‘)
cols[j] += 1;
}
if (grid[i][j] == ‘0‘ && rows + cols[j] > result)
result = rows + cols[j];
}
}
return result;
}
}
Java: Iteration
public class Solution {
public int maxKilledEnemies(char[][] A) {
if (A == null || A.length == 0 || A[0].length == 0) {
return 0;
}
int m = A.length;
int n = A[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];
int i, j, t;
for (i = 0; i < m; ++i) {
for (j = 0; j < n; ++j) {
up[i][j] = 0;
if (A[i][j] != ‘W‘) {
if (A[i][j] == ‘E‘) {
up[i][j] = 1;
}
if (i - 1 >= 0) {
up[i][j] += up[i-1][j];
}
}
}
}
for (i = m - 1; i >= 0; --i) {
for (j = 0; j < n; ++j) {
down[i][j] = 0;
if (A[i][j] != ‘W‘) {
if (A[i][j] == ‘E‘) {
down[i][j] = 1;
}
if (i + 1 < m) {
down[i][j] += down[i+1][j];
}
}
}
}
for (i = 0; i < m; ++i) {
for (j = 0; j < n; ++j) {
left[i][j] = 0;
if (A[i][j] != ‘W‘) {
if (A[i][j] == ‘E‘) {
left[i][j] = 1;
}
if (j - 1 >= 0) {
left[i][j] += left[i][j-1];
}
}
}
}
for (i = 0; i < m; ++i) {
for (j = n - 1; j >= 0; --j) {
right[i][j] = 0;
if (A[i][j] != ‘W‘) {
if (A[i][j] == ‘E‘) {
right[i][j] = 1;
}
if (j + 1 < n) {
right[i][j] += right[i][j+1];
}
}
}
}
int res = 0;
for (i = 0; i < m; ++i) {
for (j = 0; j < n; ++j) {
if (A[i][j] == ‘0‘) {
t = up[i][j] + down[i][j] + left[i][j] + right[i][j];
if (t > res) {
res = t;
}
}
}
}
return res;
}
}
Java: DP
public class Solution {
public int maxKilledEnemies(char[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int[][] enemyCount = new int[grid.length][2];
for (int i = 0; i < enemyCount.length; i++) {
enemyCount[i][0] = -1;
}
int max = 0;
for (int j = 0; j < grid[0].length; j++) {
int colEnemy = countColEnemy(grid, j, 0);
for (int i = 0; i < grid.length; i++) {
if (grid[i][j] == ‘0‘) {
if (j > enemyCount[i][0]) {
update(enemyCount, grid, i, j);
}
max = Math.max(colEnemy + enemyCount[i][1], max);
}
if (grid[i][j] == ‘W‘) {
colEnemy = countColEnemy(grid, j, i + 1);
}
}
}
return max;
}
private void update(int[][] enemyCount, char[][] grid, int i, int j) {
int count = 0;
int k = enemyCount[i][0] + 1;
while (k < grid[0].length && (grid[i][k] != ‘W‘ || k < j)) {
if (grid[i][k] == ‘E‘) {
count += 1;
}
if (grid[i][k] == ‘W‘) {
count = 0;
}
k += 1;
}
enemyCount[i][0] = k;
enemyCount[i][1] = count;
}
private int countColEnemy(char[][] grid, int j, int start) {
int count = 0;
int i = start;
while (i < grid.length && grid[i][j] != ‘W‘) {
if (grid[i][j] == ‘E‘) {
count += 1;
}
i += 1;
}
return count;
}
}