LeetCode695. Max Area of Island

最大岛屿面积求解

最新推荐文章于 2024-08-13 23:11:39 发布

原创最新推荐文章于 2024-08-13 23:11:39 发布 · 244 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#DFS

算法专栏收录该内容

19 篇文章

订阅专栏

本文介绍了一种基于深度优先搜索（DFS）算法求解二维数组中最大岛屿面积的方法。通过遍历每个元素，利用DFS算法查找相连的1（陆地），并计算其面积，最终找出最大的岛屿面积。

Description:

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]
Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]
Given the above grid, return 0.

Note:

The length of each dimension in the given grid does not exceed 50.

分析：

和逃生类似，借助前后左右四个方向（{{1,0}, {-1,0},{0,1},{0,-1}}）进行DFS。

代码：

public class Solution {

  private int area;
  private int maxArea = 0;
  private int[][] direction = {{1,0}, {-1,0},{0,1},{0,-1}};

  public int maxAreaOfIsland(int[][] grid) {
    if (null == grid || grid.length < 1){
      return 0;
    }

    for (int i = 0; i < grid.length; i++){
      for (int j = 0; j < grid[0].length; j++){
        if (grid[i][j] == 1) {
          area = 0;
          dfs(grid, i, j);
          maxArea = Math.max(maxArea, area);
        }
      }
    }

    return maxArea;
  }

  private void dfs(int[][] grid, int r, int c){
    if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length || grid[r][c] == 0){
      return ;
    }
    area++;
    grid[r][c] = 0;
    for (int i = 0; i < direction.length; i++){
      dfs(grid, r + direction[i][0], c + direction[i][1]);
    }
  }

}

题目来源：https://leetcode.com/problems/max-area-of-island/

类似题目（解题思路类似）：

1) https://leetcode.com/problems/island-perimeter/

class Solution {
    
    private int[][] direction = {{1,0}, {-1,0},{0,1},{0,-1}};
    public int islandPerimeter(int[][] grid) {
    if (grid.length < 1 || grid[0].length < 1){
      return 0;
    }
    for (int i = 0; i < grid.length; i++){
      for (int j = 0; j < grid[0].length; j++){
        if (grid[i][j] == 1){
          return dfs(grid, i, j);

        }
      }
    }
    return 0;
  }

  private int dfs(int[][] grid, int i, int j){
    if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0){
      return 1;
    }

    if (grid[i][j] == -1){
      return 0;
    }
    int cnt = 0;
    grid[i][j] = -1;
    for (int[] dir : direction){
      cnt += dfs(grid, i + dir[0], j + dir[1]);
    }

    return cnt;

  }
    
}

2) https://leetcode.com/problems/word-search/

找到每个单词的第一个匹配字母和递归以检查其余单词。为了遵守不使用位置多于一次的规则，我们设定为“0”被访问的位置。

class Solution {
  private int[][] direction = {{1,0},{-1,0},{0,1},{0,-1}};
  public boolean exist(char[][] board, String word) {

    if (board.length < 1 || board[0].length < 1 || word.trim() == ""){
      return false;
    }

    for (int i = 0; i < board.length; i++){
      for (int j = 0; j < board[0].length; j++){
        if (board[i][j] == word.charAt(0) && dfs(board, word, 0, i, j)){
          return true;
        }
      }
    }
    return false;
  }

  private boolean dfs(char[][] board,String word, int start, int r, int c){
    //当大于等于字符串长度的时候说明已经找到该字符串了
    if (word.length() <= start){
      return true;
    }
    if (r < 0 || r >= board.length || c < 0 || c >= board[0].length || board[r][c] == '0'
            || board[r][c] != word.charAt(start)){
      return false;
    }

    char tmp = board[r][c];
    board[r][c] = '0';

    for (int[] dir : direction){
      if (dfs(board, word, start+1, r+dir[0], c+dir[1])){
        return true;
      }
    }

    //还原
    board[r][c] = tmp;

    return false;
  }
}