這道題與Maze I 大同小異,不同的是需要search出最短距離。而處理方法也是不同于maze I的 visited數(shù)組,在這里采用distance數(shù)組來track每個邊界點到起點的最短距離。同時當(dāng)該點距離需要更新時,才做進(jìn)一步的搜索。
BFS:
class Solution {
public:
bool isValid(vector<vector<int>>& maze, int x, int y){
if(x >= 0 && x < maze.size() && y >= 0 && y < maze[0].size() && maze[x][y] == 0) return true;
return false;
}
int compute(int x1, int y1, int x2, int y2){
return abs(x1-x2) + abs(y1-y2);
}
int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
if(maze.empty() || maze[0].empty() || start.empty() || destination.empty()) return -1;
int row = maze.size(), col = maze[0].size();
vector<vector<int>> distance(row, vector<int>(col, INT_MAX));
distance[start[0]][start[1]] = 0;
queue<pair<int, int>> q;
q.push({start[0], start[1]});
vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while(!q.empty()){
int i = q.front().first, j = q.front().second;
q.pop();
for(auto it : directions){
int x = i, y = j;
while(isValid(maze, x + it.first, y + it.second)){
x += it.first; y += it.second;
}
if(distance[x][y] > distance[i][j] + compute(x, y, i, j)){
distance[x][y] = distance[i][j] + compute(x, y, i, j);
q.push({x, y});
}
}
}
return distance[destination[0]][destination[1]] == INT_MAX ? -1 : distance[destination[0]][destination[1]];
}
};
DFS:
class Solution {
public:
bool isValid(vector<vector<int>>& maze, int x, int y){
if(x >= 0 && x < maze.size() && y >= 0 && y < maze[0].size() && maze[x][y] == 0) return true;
return false;
}
int compute(int x1, int y1, int x2, int y2){
return abs(x1-x2) + abs(y1-y2);
}
void dfs(vector<vector<int>>& maze, vector<vector<int>> &distance, int i, int j){
int row = maze.size(), col = maze[0].size();
vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
for(auto it : directions){
int x = i, y = j;
while(isValid(maze, x + it.first, y + it.second)){
x += it.first; y += it.second;
}
if(distance[x][y] > distance[i][j] + compute(x, y, i, j)){
distance[x][y] = distance[i][j] + compute(x, y, i, j);
dfs(maze, distance, x, y);
}
}
}
int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
if(maze.empty() || maze[0].empty() || start.empty() || destination.empty()) return -1;
int row = maze.size(), col = maze[0].size();
vector<vector<int>> distance(row, vector<int>(col, INT_MAX));
distance[start[0]][start[1]] = 0;
dfs(maze, distance, start[0], start[1]);
return distance[destination[0]][destination[1]] == INT_MAX ? -1 : distance[destination[0]][destination[1]];
}
};