1905. 統計子島嶼
阿新 • • 發佈:2021-08-22
遍歷grid2,如果在遍歷一個島嶼的過程中沒有超過相應的grid1的1的範圍,說明遍歷到了一個子島嶼,否則不是子島嶼
int st[510][510]; int dx[] = {0, 1, 0, -1}; int dy[] = {1, 0, -1, 0}; int valid; class Solution { public: void dfs(int x, int y, int row, int col, vector<vector<int>>& grid1, vector<vector<int>>& grid2, int &valid){ st[x][y] = 1; if(grid1[x][y] == 0) valid = 0; for(int i = 0; i < 4; i ++){ int a = x + dx[i], b = y + dy[i]; if(a < 0 || a >= row || b < 0 || b >= col) continue; if(grid2[a][b] && !st[a][b]) dfs(a, b, row, col, grid1, grid2, valid); } } int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) { memset(st, 0, sizeof st); int row = grid1.size(), col = grid1[0].size(); int cnt = 0; for(int i = 0; i < row; i ++){ for(int j = 0; j < col; j ++){ if(grid2[i][j] && st[i][j] == 0){ valid = 1; dfs(i, j, row, col, grid1, grid2, valid); cnt += valid; } } } return cnt; } };