LeetCode695. Max Area of Island_leetcode max area of island c++-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_31663969/article/details/79048243

本文介绍了一种使用深度优先搜索（DFS）算法来解决寻找二维数组中最大岛屿面积的问题。通过遍历每个岛屿并标记已访问区域，确保了正确计算岛屿的最大面积。

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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.

题目分析：

题目意思是给定一个方格，其中1表示地面，0表示水，要我们求出其中最大的“小岛”的面积

采用dfs的思路遍历所有相邻的1，每遍历到一个面积+1，直到不能遍历到1为止，然后返回这个小岛的面积和之前得到的做比较，大的保留。

c++代码：

class Solution {
public:
    int max(int a, int b) {
        if (a > b) return a;
        else return b;
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int res = 0;
        for(int i = 0; i < grid.size(); i++){
            for(int j = 0; j < grid[0].size(); j++){
                if(grid[i][j] == 1) res = max(res, helper(grid, i, j));
            }
        }
        return res;
    }
    int helper(vector<vector<int>>& grid, int r, int c){
        if(r < 0 || r >= grid.size() || c < 0 || c >= grid[0].size()) return 0;
        int res = 0;
        if(grid[r][c] == 1) {
            grid[r][c] = 0;
            res = 1 + helper(grid, r-1, c) + helper(grid, r+1, c) + helper(grid, r, c-1) + helper(grid, r, c+1); 
        }
        return res;
    }
};