【启发】leetcode - 695. Max Area of Island【回溯法 + 图的遍历】

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本文介绍了一种通过深度优先搜索算法解决二维数组中寻找最大岛屿面积的问题。利用递归回溯的方法，对每个可能成为岛屿起点的位置进行遍历，并通过标记已访问节点避免重复计算，最终找出最大岛屿面积。

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题目

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.

题意

给定一个二维数组，计算其中最大岛屿的面积。（注意形成岛屿的规则）

分析及解答

【遍历法】下面的解法，将问题当作图来进行求解（使用了visiteds数组用来记录访问状态）。
【实现方式】深度优先遍历，回溯法。

class Solution {
    
	public int maxAreaOfIsland(int[][] grid) {
		if (grid == null || grid.length == 0) {
			return 0;
		}
		
		int[][] visited = new int[grid.length][grid[0].length];
		int max = 0;
		for (int i = 0; i < grid.length; i++) {
			for (int j = 0; j < grid[0].length; j++) {

				if (grid[i][j] == 0) {
					continue;
				}
				int result = find(grid,i,j,visited);
				if(max < result){
					max = result;
				}
			}
		}
		return max;
	}
public int find(int[][] grid ,int x,int y, int[][] visited){
		if(grid[x][y] == 0 || visited[x][y] == 1){ //边界溢出。
			return 0;
		}
		int result = 1;
		visited[x][y] = 1;
		if(x + 1 < grid.length){
			result += find(grid, x+1, y, visited);
		}
		if(x -1 >= 0){
			result += find(grid, x-1, y, visited);
		}
		if(y + 1 < grid[0].length){
			result += find(grid,x,y+1,visited);
		}
		if(y -1 >= 0){
			result += find(grid,x,y-1,visited);
		}
		return result;
	}
}