My code:
public class Solution {
public int maxKilledEnemies(char[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int max = 0;
int row = 0;
int[] cols = new int[grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (j == 0 || grid[i][j - 1] == 'W') {
row = updateRow(i, j, grid);
}
if (i == 0 || grid[i - 1][j] == 'W') {
cols[j] = updateCol(i, j, grid);
}
if (grid[i][j] == '0') {
max = Math.max(max, row + cols[j]);
}
}
}
return max;
}
private int updateRow(int i, int j, char[][] grid) {
int counter = 0;
for (int k = j; k < grid[0].length; k++) {
if (grid[i][k] == 'W') {
break;
}
else if (grid[i][k] == 'E') {
counter++;
}
}
return counter;
}
private int updateCol(int i, int j, char[][] grid) {
int counter = 0;
for (int k = i; k < grid.length; k++) {
if (grid[k][j] == 'W') {
break;
}
else if (grid[k][j] == 'E') {
counter++;
}
}
return counter;
}
}
reference:
https://discuss.leetcode.com/topic/48742/simple-dp-solution-in-java/2
这道题目我自己也做了出来。
首先是 brute-force,
然后思考,哪里可以加cache
发现,从列角度看,如果左侧是’0′,那么,当前这个点,这一行的炸死敌人数,就等于左边这一点,这一行的数。
同理,从行角度看,如果上侧是’0′,那么,当前这个点,这一列的炸死敌人数,就等于上边这一点,这一列的数。
用这个原理加cache
因为每个点都有 行炸死个数,列炸死个数。
我搞了两个二维数组作为cache
int[][] rows
int[][] cols
其实原理和上面的解法原理是相同的。只是cache规模太大,浪费了很多时间。
我的解法如下:
My code:
public class Solution {
public int maxKilledEnemies(char[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int max = 0;
int[][] cols = new int[grid.length][grid[0].length];
int[][] rows = new int[grid.length][grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '0') {
max = Math.max(max, place(i, j, grid, rows, cols));
}
}
}
return max;
}
private int place(int i, int j, char[][] grid, int[][] rows, int[][] cols) {
int row = 0;
int col = 0;
boolean colHit = false;
boolean rowHit = false;
if (i > 0 && grid[i - 1][j] == '0') {
col += cols[i - 1][j];
colHit = true;
}
if (j > 0 && grid[i][j - 1] == '0') {
row += rows[i][j - 1];
rowHit = true;
}
if (!colHit) {
// up
for (int k = i - 1; k >= 0; k--) {
if (grid[k][j] == 'W') {
break;
}
else if (grid[k][j] == 'E') {
col++;
}
}
// down
for (int k = i + 1; k < grid.length; k++) {
if (grid[k][j] == 'W') {
break;
}
else if (grid[k][j] == 'E') {
col++;
}
}
}
if (!rowHit) {
// left
for (int k = j - 1; k >= 0; k--) {
if (grid[i][k] == 'W') {
break;
}
else if (grid[i][k] == 'E') {
row++;
}
}
// right
for (int k = j + 1; k < grid[0].length; k++) {
if (grid[i][k] == 'W') {
break;
}
else if (grid[i][k] == 'E') {
row++;
}
}
}
cols[i][j] = col;
rows[i][j] = row;
return col + row;
}
}
Anyway, Good luck, Richardo! — 09/20/2016