695.Max-Area-of-Island

695. Max Area of Island

题目地址

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

题目描述

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.

代码

Approach #1 DFS

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
      int R = grid.length;
      int C = grid[0].length;
        boolean seen = new boolean[R][C];
      int ans = 0;
      for (int r = 0; r < R; r++) {
        for (int c = 0; c < C; c++) {
          ans = Math.max(ans, area(r, c, seen, grid));
        }
      }
      return ans;
    }

  private int area(int r, int c, boolean[][] seen, int[][] grid) {
    if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length
       || seen[r][c] || grid[r][c] == 0) {
      return 0;
    }
    seen[r][c] = true;
    return (1 + area(r + 1, c, seen, grid) 
            + area(r - 1, c, seen, grid)
            + area(r, c - 1, seen, grid) 
            + area(r, c + 1, seen, grid));
  }

}

Approach #2 DFS Iterative

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        boolean[][] seen = new boolean[grid.length][grid[0].length];
        int[] dr = new int[]{1, -1, 0, 0};
        int[] dc = new int[]{0, 0, 1, -1};

        int ans = 0;
        for (int r0 = 0; r0 < grid.length; r0++) {
            for (int c0 = 0; c0 < grid[0].length; c0++) {
                if (grid[r0][c0] == 1 && !seen[r0][c0]) {

                    int shape = 0;
                    Stack<int[]> stack = new Stack();
                    stack.push(new int[]{r0, c0});
                    seen[r0][c0] = true;
                    while (!stack.empty()) {
                        int[] node = stack.pop();
                        int r = node[0], c = node[1];
                        shape++;
                        for (int k = 0; k < 4; k++) {
                            int nr = r + dr[k];
                            int nc = c + dc[k];
                            if (0 <= nr && nr < grid.length &&
                                    0 <= nc && nc < grid[0].length &&
                                    grid[nr][nc] == 1 && !seen[nr][nc]) {
                                stack.push(new int[]{nr, nc});
                                seen[nr][nc] = true;
                            }
                        }
                    }
                    ans = Math.max(ans, shape);
                }
            }
        }
        return ans;
    }
}