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我试过下面的代码，但它没有给我正确的答案。如何在钥匙和门的拼图中找到最短路径
这是问题陈述。
假设你有一个2-D网格。每个点都是土地或水。 也是一个起点和目标。
现在有钥匙可以打开门。每个钥匙对应一个 门。
实现，使用土地的瓷砖，钥匙开门返回 的目标从一开始的最短路径的功能。
数据表示法
该映射将作为字符串数组传递。
一张地图可以有以下图块。
0 = Water1 = Land2 = Start3 = Goaluppercase = doorlowercase = key网格可以是这样的：
{{'0', '2', '1', '1', '1'}, {'0', '1', '0', '0', '1'}, {'0', '0', '0', '0', '1'}, {'0', '0', 'A', '0', '1'}, {'1', '1', 'a', '1', '1'}, {'1', 'b', '0', '0', 'B'}, {'1', '1', '0', '0', '1'}, {'0', '1', '0', '0', '3'}};所以最短路径应该是： 01-> 02-> 03-> 04-> 14 →24→34→44→43→42→41→51→41→42→43→44→54→64→74
我的代码去像：
public class shortestPath {
static int shortest = Integer.MAX_VALUE;
static String path = "";
static ArrayList<ArrayList<int[]>> res = new ArrayList<ArrayList<int[]>>();
public static void main(String[] args) {
char[][] board = {{'0', '2', '1', '1', '1'},
{'0', '1', '0', '0', '1'},
{'0', '0', '0', '0', '1'},
{'0', '0', 'A', '0', '1'},
{'1', '1', 'a', '1', '1'},
{'1', 'b', '0', '0', 'B'},
{'1', '1', '0', '0', '1'},
{'0', '1', '0', '0', '3'}};
System.out.println(findShortest(board));
System.out.println(path);
}
public static int findShortest(char[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) return 0;
int row = board.length;
int col = board[0].length;
int[] start = new int[2];
int[] end = new int[2];
int[][] visited = new int[row][col];
HashMap<Character, ArrayList<int[]>> hm = new HashMap<>();
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (board[i][j] == '2') {
start = new int[]{i, j};
}
if (board[i][j] == '3') {
end = new int[]{i, j};
}
if (Character.isUpperCase(board[i][j])) {
char key = Character.toLowerCase(board[i][j]);
if (hm.containsKey(key)) {
hm.get(key).add(new int[]{i, j});
} else {
ArrayList<int[]> al = new ArrayList<>();
al.add(new int[]{i, j});
hm.put(key, al);
}
board[i][j] = '0';
}
}
}
//HashSet<Character> hs = new HashSet<Character>();
//helper2(board, hs, start[0], start[1], row, col, 0, visited, new StringBuilder(), new ArrayList<int[]>(), res);
helper(board, hm, start[0], start[1], row, col, 0, visited, new StringBuilder());
return shortest;
}
public static void helper(char[][] board, HashMap<Character, ArrayList<int[]>> hm, int r, int c, int row, int col, int count, int[][] visited, StringBuilder sb) {
// System.out.println(r + " " + c);
visited[r][c] = visited[r][c] + 1;
sb.append(r).append(c).append("->");
if (board[r][c] == '3') {
System.out.println(count);
if (shortest > count) {
path = sb.deleteCharAt(sb.length() - 1).deleteCharAt(sb.length() - 1).toString();
sb.append("->");
shortest = count;
}
//return;
//shortest = Math.min(shortest, count);
}
if (Character.isLowerCase(board[r][c])) { // if this is the key;
if (hm.containsKey(board[r][c])) {
for (int[] i : hm.get(board[r][c])) {
board[i[0]][i[1]] = '1';
}
}
}
if (r + 1 < row && board[r + 1][c] != '0' && visited[r + 1][c] < 2) {
helper(board, hm, r + 1, c, row, col, count + 1, visited, sb);
}
if (c + 1 < col && board[r][c + 1] != '0' && visited[r][c + 1] < 2) {
helper(board, hm, r, c + 1, row, col, count + 1, visited, sb);
}
if (r - 1 >= 0 && board[r - 1][c] != '0' && visited[r - 1][c] < 2) {
helper(board, hm, r - 1, c, row, col, count + 1, visited, sb);
}
if (c - 1 >= 0 && board[r][c - 1] != '0' && visited[r][c - 1] < 2) {
helper(board, hm, r, c - 1, row, col, count + 1, visited, sb);
}
visited[r][c] = visited[r][c] - 1;
sb.delete(sb.length() - 4, sb.length());
if (Character.isLowerCase(board[r][c])) { // if this is the key;
if (hm.containsKey(board[r][c])) {
for (int[] i : hm.get(board[r][c])) {
board[i[0]][i[1]] = '0';
}
}
}
return;
}
}
你做了什么**你得到答案？你是否已经通过调试器完成了你的算法？ –
使用调试器确实是解决此问题的唯一好方法。我怀疑有太多的代码可以让任何人通过阅读来发现错误。 –