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200、岛屿数量

• 🔗：200. 岛屿数量 - 力扣（LeetCode）
• 思路：
  • 1. 深度优先算法
    • 二叉树中dfs要素：1、访问左右相邻子节点 2、判断base case（终止条件）
    • 参考二叉树中的dfs看网格问题
    • 1. 网格的相邻节点：上下左右4个
    • 2.终止条件：超出格子的范围（--对应二叉树中全部为null的base case）
    • 3. 关键！！避免重复遍历，做过的格子需要进行标记
  • 2. 广度优先算法
    • 扫描整个二维网格，遇到为1的格子，加入队列当中，进行广度搜索
• 代码
  • 深度优先算法

  • class Solution {public int numIslands(char[][] grid) {int area = 0;for(int i=0; i<grid.length; i++){for(int j=0; j<grid[0].length; j++){if(grid[i][j] == '1'){area++;dfs(grid, i, j);}}}return area;}private void dfs(char[][] grid, int r, int c){if(!isArea(grid,r,c)){return;}if(grid[r][c] != '1'){return;}grid[r][c] = '2';dfs(grid, r-1, c);dfs(grid, r, c-1);dfs(grid, r+1, c);dfs(grid, r, c+1);}boolean isArea(char[][] grid, int r, int c){return 0<=r && r < grid.length && 0 <= c && c < grid[0].length;}
    }

  • 广度优先算法

    • class Solution {/**广度优先搜索bfs扫描整个二维网络，如果一个位置为1，加入队列，进行广度优先搜索*/public int numIslands(char[][] grid) {if(grid == null || grid.length == 0){return 0;}int nr = grid.length;int nc = grid[0].length;int nums_islands = 0;for(int r=0; r < nr; ++r){for(int c = 0; c<nc; ++c){if(grid[r][c] == '1'){++nums_islands;grid[r][c] = '2';Queue<Integer> neighbors = new LinkedList<>();neighbors.add(r * nc + c);while(!neighbors.isEmpty()){int id = neighbors.remove();int row = id / nc;int col = id % nc;if (row - 1 >= 0 && grid[row-1][col] == '1') {grid[row-1][col] = '2';neighbors.add((row-1) * nc + col);}if (row + 1 < nr && grid[row+1][col] == '1') {grid[row+1][col] = '2';neighbors.add((row+1) * nc + col);}if (col - 1 >= 0 && grid[row][col-1] == '1') {neighbors.add(row * nc + col-1);grid[row][col-1] = '2';}if (col + 1 < nc && grid[row][col+1] == '1') {neighbors.add(row * nc + col+1);grid[row][col+1] = '2';}                      }}}}return nums_islands;}

       

695. 岛屿的最大面积

• 🔗：695. 岛屿的最大面积 - 力扣（LeetCode）
• 思路：深度优先搜索
• 代码：

  class Solution {public int maxAreaOfIsland(int[][] grid) {if(grid.length==0||grid[0].length==0){return 0;}int res = 0;for(int r=0; r<grid.length; r++){for(int c=0; c<grid[0].length; c++){if(grid[r][c]==1){int a = area(grid, r, c);res = Math.max(res,a);}}}return res;}int area(int[][] grid, int r, int c){if (!(0 <= r && r < grid.length && 0 <= c && c < grid[0].length)) {return 0;}if(grid[r][c] != 1){return 0;}grid[r][c] = 2;return 1 + area(grid, r-1, c)+ area(grid, r+1, c)+ area(grid, r, c-1)+ area(grid, r, c+1);}
  }