You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map:
0 represents the obstacle can't be reached.
1 represents the ground can be walked through.
The place with number bigger than 1 represents a tree can be walked through, and this positive number represents the tree's height.
You are asked to cut off all the trees in this forest in the order of tree's height - always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).
You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the trees. If you can't cut off all the trees, output -1 in that situation.
You are guaranteed that no twotreeshave the same height and there is at least one tree needs to be cut off.
Example 1:
Input:
[
[1,2,3],
[0,0,4],
[7,6,5]
]
Output:
6
Example 2:
Input:
[
[1,2,3],
[0,0,0],
[7,6,5]
]
Output:
-1
Example 3:
Input:
[
[2,3,4],
[0,0,5],
[8,7,6]
]
Output:
6
Explanation:
You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking.
Hint: size of the given matrix will not exceed 50x50.
class Solution {
public int cutOffTree(List<List<Integer>> forest) {
// Get all the trees with h-height, r-row, c-column and sort by height
List<int[]> trees = new ArrayList<>();
int[][] map = new int[forest.size()][forest.get(0).size()];
for (int r = 0; r < forest.size(); r++) {
for (int c = 0; c < forest.get(r).size(); c++) {
int h = forest.get(r).get(c);
map[r][c] = h;
if (h > 1) {
trees.add(new int[] {h, r, c});
}
}
}
// Sort trees based on height of trees
Collections.sort(trees, (t1, t2) -> Integer.compare(t1[0], t2[0]));
int ans = 0;
int r0 = 0;
int c0 = 0;
// Find distance from current position to next tree
for (int[] tree: trees) {
int d = dist(map, r0, c0, tree[1], tree[2]);
if (d < 0) {
return -1;
}
ans += d;
r0 = tree[1];
c0 = tree[2];
}
return ans;
}
private static int[][] dirs = {{-1, 0}, {1, 0} , {0, -1}, {0, 1}};
private int dist(int[][] map, int sr, int sc, int tr, int tc) {
int distance = 0;
boolean[][] visited = new boolean[map.length][map[0].length];
Queue<int[]> q = new LinkedList<int[]>();
q.offer(new int[] {sr, sc});
visited[sr][sc] = true;
while (!q.isEmpty()) {
int levelSize = q.size();
while (levelSize-- > 0) {
int[] pos = q.poll();
if (pos[0] == tr && pos[1] == tc) {
return distance;
}
for (int[] dir: dirs) {
int nr = pos[0] + dir[0];
int nc = pos[1] + dir[1];
if (nr >= 0 && nr < map.length && nc >= 0 && nc < map[0].length) {
if (map[nr][nc] != 0 && !visited[nr][nc]) {
q.offer(new int[] {nr, nc});
visited[nr][nc] = true;
}
}
}
}
distance++;
}
return -1;
}
}
Another BFS (Store distance in int[] {r, c, dist})
类似上面的一层一层BFS，这里就是把距离放到每个节点的int[]中.
Runtime: 258 ms, faster than 84.21% of Java online submissions for Cut Off Trees for Golf Event.
Memory Usage: 46.7 MB, less than 100.00% of Java online submissions for Cut Off Trees for Golf Event.
class Solution {
public int cutOffTree(List<List<Integer>> forest) {
// Get all the trees with h-height, r-row, c-column and sort by height
List<int[]> trees = new ArrayList<>();
int[][] map = new int[forest.size()][forest.get(0).size()];
for (int r = 0; r < forest.size(); r++) {
for (int c = 0; c < forest.get(r).size(); c++) {
int h = forest.get(r).get(c);
map[r][c] = h;
if (h > 1) {
trees.add(new int[] {h, r, c});
}
}
}
// Sort trees based on height of trees
Collections.sort(trees, (t1, t2) -> Integer.compare(t1[0], t2[0]));
int ans = 0;
int r0 = 0;
int c0 = 0;
// Find distance from current position to next tree
for (int[] tree: trees) {
int d = dist(forest, r0, c0, tree[1], tree[2]);
if (d < 0) {
return -1;
}
ans += d;
r0 = tree[1];
c0 = tree[2];
}
return ans;
}
private static int[][] dirs = {{-1, 0}, {1, 0} , {0, -1}, {0, 1}};
public int dist(List<List<Integer>> forest, int sr, int sc, int tr, int tc) {
int R = forest.size(), C = forest.get(0).size();
Queue<int[]> queue = new LinkedList();
boolean[][] seen = new boolean[R][C];
queue.add(new int[]{sr, sc, 0});
seen[sr][sc] = true;
while (!queue.isEmpty()) {
int[] cur = queue.poll();
if (cur[0] == tr && cur[1] == tc) {
return cur[2];
}
for (int[] dir: dirs) {
int r = cur[0] + dir[0];
int c = cur[1] + dir[1];
if (0 <= r && r < R && 0 <= c && c < C &&
!seen[r][c] && forest.get(r).get(c) > 0) {
seen[r][c] = true;
queue.add(new int[]{r, c, cur[2] + 1});
}
}
}
return -1;
}
}
Runtime: 911 ms, faster than 3.64% of Java online submissions for Cut Off Trees for Golf Event.
MemoryUsage: 46.7 MB, less than 100.00% of Java online submissions for Cut Off Trees for Golf Event.
class Solution {
public int cutOffTree(List<List<Integer>> forest) {
// Get all the trees with h-height, r-row, c-column and sort by height
List<int[]> trees = new ArrayList<>();
int[][] map = new int[forest.size()][forest.get(0).size()];
for (int r = 0; r < forest.size(); r++) {
for (int c = 0; c < forest.get(r).size(); c++) {
int h = forest.get(r).get(c);
map[r][c] = h;
if (h > 1) {
trees.add(new int[] {h, r, c});
}
}
}
// Sort trees based on height of trees
Collections.sort(trees, (t1, t2) -> Integer.compare(t1[0], t2[0]));
int ans = 0;
int r0 = 0;
int c0 = 0;
// Find distance from current position to next tree
for (int[] tree: trees) {
int d = dist(map, r0, c0, tree[1], tree[2]);
if (d < 0) {
return -1;
}
ans += d;
r0 = tree[1];
c0 = tree[2];
}
return ans;
}
private static int[][] dirs = {{-1, 0}, {1, 0} , {0, -1}, {0, 1}};
// Dijkstra's Algorithm to Find Shortest Distance
private int dist(int[][] map, int sr, int sc, int tr, int tc) {
// Init distance matrix
int[][] distance = new int[map.length][map[0].length];
for (int[] row: distance) {
Arrays.fill(row, Integer.MAX_VALUE);
}
distance[sr][sc] = 0;
// Init priority queue for vertices to traverse
PriorityQueue<int[]> q = new PriorityQueue<int[]>((a, b) -> a[2] - b[2]);
q.offer(new int[] {sr, sc, 0});
while (!q.isEmpty()) {
int[] pos = q.poll();
for (int[] dir: dirs) {
int nr = pos[0] + dir[0];
int nc = pos[1] + dir[1];
if (nr >= 0 && nr < map.length && nc >= 0 && nc < map[0].length) {
// Update distance matrix and add the vertex to priority queue
if (map[nr][nc] != 0 && (pos[2] + 1 < distance[nr][nc])) {
distance[nr][nc] = pos[2] + 1;
q.offer(new int[] {nr, nc, pos[2] + 1});
}
}
}
}
if (distance[tr][tc] == Integer.MAX_VALUE) {
return -1;
}
return distance[tr][tc];
}
}