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最新推荐文章于 2025-06-11 18:07:26 发布

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最新推荐文章于 2025-06-11 18:07:26 发布

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分类专栏： leetcode刷题记录文章标签： leetcode

原文链接：https://leetcode-cn.com/problems/number-of-islands/solution/number-of-islands-shen-du-you-xian-bian-li-dfs-or-/

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思路一：深度优先遍历DFS

class Solution {
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == '1'){
                    dfs(grid, i, j);
                    count++;
                }
            }
        }
        return count;
    }
    private void dfs(char[][] grid, int i, int j){
        if(i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') return;
        grid[i][j] = '0';
        dfs(grid, i + 1, j);
        dfs(grid, i, j + 1);
        dfs(grid, i - 1, j);
        dfs(grid, i, j - 1);
    }
}

思路二：广度优先遍历 BFS

class Solution {
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == '1'){
                    bfs(grid, i, j);
                    count++;
                }
            }
        }
        return count;
    }
    private void bfs(char[][] grid, int i, int j){
        Queue<int[]> list = new LinkedList<>();
        list.add(new int[] { i, j });
        while(!list.isEmpty()){
            int[] cur = list.remove();
            i = cur[0]; j = cur[1];
            if(0 <= i && i < grid.length && 0 <= j && j < grid[0].length && grid[i][j] == '1') {
                grid[i][j] = '0';
                list.add(new int[] { i + 1, j });
                list.add(new int[] { i - 1, j });
                list.add(new int[] { i, j + 1 });
                list.add(new int[] { i, j - 1 });
            }
        }
    }
}