zoukankan      html  css  js  c++  java
  • leetcode 695. Max Area of Island

    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.

    思路:计算每个岛屿的面积。(深搜)

     1 class Solution {
     2 public:
     3     int maxAreaOfIsland(vector<vector<int>>& grid) {
     4         int maxArea = 0;
     5         int m = grid.size();
     6         if (m == 0)
     7             return 0;
     8         int n = grid[0].size();
     9         for (int i = 0; i < m; i++) {
    10             for (int j = 0; j < n; j++) {
    11                 if (grid[i][j] == 1) {
    12                     int tmp = dfs(i, j, grid);
    13                     maxArea = max(maxArea, tmp);
    14                 }
    15             }
    16         }
    17         return maxArea;
    18     }
    19 private:
    20     int dfs(int i, int j, vector<vector<int> > &grid) {
    21         grid[i][j] = 0;
    22         int cnt = 1;
    23         int dx[4] = {-1, 0, 1, 0};
    24         int dy[4] = {0, 1, 0, -1};
    25         //four dimension
    26         for (int d = 0; d < 4; d++) {
    27             int newx = i + dx[d];
    28             int newy = j + dy[d];
    29             if (newx >= 0 && newx < grid.size() && newy >= 0 && newy < grid[0].size() && grid[newx][newy] == 1) {
    30                 cnt += dfs(newx, newy, grid);
    31             }
    32         }
    33         return cnt;
    34     }
    35 };
  • 相关阅读:
    Generate Parentheses (Java)
    leetcode15
    MD5
    leetcode409
    Vue第一个简单的例子
    SpringBoot和Ajax通信
    如何使用安装光盘为本机创建yum repository
    Less known Solaris features: svccfg editprop (ZT)
    Rename Oracle Managed File (OMF) datafiles in ASM(ZT)
    跨数据文件删除flashback database
  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/11551112.html
Copyright © 2011-2022 走看看