LC 959. Regions Cut By Slashes

In a N x N grid composed of 1 x 1 squares, each 1 x 1 square consists of a /, \, or blank space.  These characters divide the square into contiguous regions.

(Note that backslash characters are escaped, so a \ is represented as "\\".)

Return the number of regions.

我的DFS解法。

Runtime: 32 ms, faster than 12.39% of C++ online submissions for Regions Cut By Slashes.

#include <string>
#include <iostream>
#include <vector>
using namespace std;
int dirs[][] = {{,},{,-},{,},{-,},{,},{,-},{-,},{-,-}};
class Solution {
public:
  int regionsBySlashes(vector<string>& grid) {
    size_t n = grid.size();
    vector<vector<int>> mtx(*n, vector<int>(*n, ));
    for(int i=; i<n; i++){
      for(int j=; j<n; j++){
        if(grid[i][j] == '/'){
          mtx[*i][*j+] = ;
          mtx[*i+][*j+] = ;
          mtx[*i+][*j+] = ;
          mtx[*i+][*j] = ;
        }else if(grid[i][j] == '\\'){
          mtx[*i][*j] = ;
          mtx[*i+][*j+] = ;
          mtx[*i+][*j+] = ;
          mtx[*i+][*j+] = ;
        }
      }
    }
    int ret = ;
    for(int i=; i<*n; i++){
      for(int j=; j<*n; j++){
        if(mtx[i][j] == ) ret++;
        dfs(mtx, i, j);
      }
    }
    return ret;
  }
  void dfs(vector<vector<int>>& mtx, int x, int y){
    size_t n = mtx.size();
    if(x <  || x >= n || y <  || y >= n || mtx[x][y] != ) return ;
    mtx[x][y] = -;
    for(int i=; i<; i++){
      int newx = x+dirs[i][];
      int newy = y+dirs[i][];
      if(newx >=  && newy >=  && newx < n && newy < n && i >= ){
        if(mtx[newx][y] ==  && mtx[x][newy] == ) continue;
      }
      dfs(mtx, newx, newy);
    }
  }
};

网上的unionfound解法。

class Solution {
public:
   int count, n;
   vector<int> f;
   int regionsBySlashes(vector<string>& grid) {
        n = grid.size();
        count = n * n * ;
        for (int i = ; i < n * n * ; ++i)
            f.push_back(i);
        for (int i = ; i < n; ++i) {
            for (int j = ; j < n; ++j) {
                if (i > ) uni(g(i - , j, ), g(i, j, ));
                if (j > ) uni(g(i , j - , ), g(i , j, ));
                if (grid[i][j] != '/') {
                    uni(g(i , j, ), g(i , j,  ));
                    uni(g(i , j, ), g(i , j,  ));
                }
                if (grid[i][j] != '\\') {
                    uni(g(i , j, ), g(i , j,  ));
                    uni(g(i , j, ), g(i , j,  ));
                }
            }
        }
        return count;
    }

    int find(int x) {
        if (x != f[x]) {
            f[x] = find(f[x]);
        }
        return f[x];
    }
    void uni(int x, int y) {
        x = find(x); y = find(y);
        if (x != y) {
            f[x] = y;
            count--;
        }
    }
    int g(int i, int j, int k) {
        return (i * n + j) *  + k;
    }
};