[LeetCode] Algorithms-695. Max Area of Island

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.

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

思路

就是一个深度优先的问题。首先遍历整个矩阵，当遍历到的元素为1时，然后以此元素为起点进行深度优先搜索，如果搜索到的元素为1时，再以次元素继续进行深度优先搜索，直到所有的条件都不满足，得到最大值。

代码

class Solution {public:    int maxAreaOfIsland(vector<vector<int>>& grid) {        int length=0;        for(int i = 0; i < grid.size(); i++) {            for(int j = 0; j < grid[0].size(); j++) {                if(grid[i][j] == 1) {                    length = max(length, maxLenghth(grid, i, j));                }            }        }        return length;      }    int maxLenghth(vector<vector<int>>& grid, int i, int j) {        if(i >= 0 && i < grid.size() && j >= 0 && j < grid[0].size() && grid[i][j] == 1) {            grid[i][j] = 0;            return 1+maxLenghth(grid, i-1, j)+maxLenghth(grid, i+1, j)+maxLenghth(grid, i, j-1)+maxLenghth(grid, i, j+1);        }        return 0;    }};