※ 引述 《argorok (死肥肥社管)》 之銘言： : 　 : 1905. Count Sub Islands : 　 : 給兩個grid 求grid2的islands中是grid1的sub island的數量 sub island : 當前island 全部的格子都在grid1的那個island裡面 只要超出去一個就不算 思路 : 一邊遍歷所有grid2的island 順便看當前island 有沒有全部跟grid1的island 重疊 然後加起來就好了 ```cpp class Solution { public: vector<vector<int>> paper; void go(vector<vector<int>>& grid , int i , int j , int &jiwp) { if(i<0 || i>=grid.size() || j<0 || j>=grid[0].size())return; if(grid[i][j]==0)return; grid[i][j]=0; if(paper[i][j]==0)jiwp = 0; go(grid,i-1,j,jiwp); go(grid,i+1,j,jiwp); go(grid,i,j-1,jiwp); go(grid,i,j+1,jiwp); } int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) { paper = grid1; int n = grid2.size(); int m = grid2[0].size(); int res = 0; for(int i = 0 ; i < n ; i ++) { for(int j = 0 ; j < m ; j++) { if(grid2[i][j]==0)continue; int jiwp = 1; go(grid2,i,j,jiwp); if(jiwp)res++; } } return res; } }; ``` -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 1.162.18.201 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/Marginalman/M.1724815859.A.E4D.html