zoukankan      html  css  js  c++  java
  • Number of Islands——LeetCode

    Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

    Example 1:

    11110
    11010
    11000
    00000

    Answer: 1

    Example 2:

    11000
    11000
    00100
    00011

    Answer: 3

    哈哈,这是今天的新题,十多分钟A了,题目不难,我的做法是BFS,题目大意是1代表岛屿,0代表水,可以假设这个grid周围都是水,岛屿定义是四周都是水,求岛屿的个数。

    我的做法是,从第一个开始,如果是1,并且没有访问过visit为false,那么加入queue里,当queue非空取出来并加入它上下左右的邻居,并且把这些节点visit设置为true,时间复杂度是O(M*N),空间复杂度的话,因为同时在queue里的最多也就4个节点的坐标,所以是O(1)。

    Talk is cheap>>

    public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0) {
                return 0;
            }
            int rowLen = grid.length;
            int colLen = grid[0].length;
            boolean[][] visited = new boolean[rowLen][colLen];
            int[][] direct = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            Deque<Integer> queue = new ArrayDeque<>();
            int res = 0;
            for (int i = 0; i < rowLen; i++) {
                for (int j = 0; j < colLen; j++) {
                    if (!visited[i][j] && grid[i][j] == '1') {
                        res++;
                        queue.add(i * colLen + j);
                        while (!queue.isEmpty()) {
                            int pos = queue.poll();
                            int curr_x = pos / colLen;
                            int curr_y = pos % colLen;
                            if (visited[curr_x][curr_y]) {
                                continue;
                            }
                            visited[curr_x][curr_y] = true;
                            for (int k = 0; k < 4; k++) {
                                int x = curr_x + direct[k][0];
                                int y = curr_y + direct[k][1];
                                if (x >= 0 && y >= 0 && x < rowLen && y < colLen && grid[x][y] == '1') {
                                    queue.add(x * colLen + y);
                                }
                            }
                        }
                    }
                }
            }
            return res;
        }

  • 相关阅读:
    Android AHandle AMessage
    android java 与C 通过 JNI双向通信
    android 系统给应用的jar
    UE4 unreliable 同步问题
    UE4 difference between servertravel and openlevel(多人游戏的关卡切换)
    UE4 Run On owing Client解析(RPC测试)
    UE4 TSubclassOf VS Native Pointer
    UE4 内容示例网络同步Learn
    UE4 多人FPS VR游戏制作笔记
    UE4 分层材质 Layerd Materials
  • 原文地址:https://www.cnblogs.com/aboutblank/p/4405557.html
Copyright © 2011-2022 走看看