企业官网建设流程全解析

岛屿问题的核心是 DFS 遍历二维数组，把连通的 1 标记为已访问。

一、岛屿数量

publicintnumIslands(char[][]grid){intcount=0;for(inti=0;i<grid.length;i++){for(intj=0;j<grid[0].length;j++){if(grid[i][j]=='1'){dfs(grid,i,j);count++;}}}returncount;}privatevoiddfs(char[][]grid,intr,intc){if(r<0||c<0||r>=grid.length||c>=grid[0].length||grid[r][c]!='1')return;grid[r][c]='0';// 标记已访问dfs(grid,r-1,c);dfs(grid,r+1,c);dfs(grid,r,c-1);dfs(grid,r,c+1);}

二、岛屿的最大面积

publicintmaxAreaOfIsland(int[][]grid){intmaxArea=0;for(inti=0;i<grid.length;i++){for(intj=0;j<grid[0].length;j++){if(grid[i][j]==1){maxArea=Math.max(maxArea,dfs(grid,i,j));}}}returnmaxArea;}privateintdfs(int[][]grid,intr,intc){if(r<0||c<0||r>=grid.length||c>=grid[0].length||grid[r][c]!=1)return0;grid[r][c]=0;return1+dfs(grid,r-1,c)+dfs(grid,r+1,c)+dfs(grid,r,c-1)+dfs(grid,r,c+1);}

三、岛屿的周长

publicintislandPerimeter(int[][]grid){for(inti=0;i<grid.length;i++){for(intj=0;j<grid[0].length;j++){if(grid[i][j]==1)returndfs(grid,i,j);}}return0;}privateintdfs(int[][]grid,intr,intc){if(r<0||c<0||r>=grid.length||c>=grid[0].length||grid[r][c]==0)return1;if(grid[r][c]==2)return0;grid[r][c]=2;returndfs(grid,r-1,c)+dfs(grid,r+1,c)+dfs(grid,r,c-1)+dfs(grid,r,c+1);}

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