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

    A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. 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:

    Given m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]].
    Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).

    0 0 0
    0 0 0
    0 0 0
    

    Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land.

    1 0 0
    0 0 0   Number of islands = 1
    0 0 0
    

    Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land.

    1 1 0
    0 0 0   Number of islands = 1
    0 0 0
    

    Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land.

    1 1 0
    0 0 1   Number of islands = 2
    0 0 0
    

    Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land.

    1 1 0
    0 0 1   Number of islands = 3
    0 1 0
    

    We return the result as an array: [1, 1, 2, 3]

    Challenge:

    Can you do it in time complexity O(k log mn), where k is the length of the positions?

    Solution:

    the solution is similar to "Number of Connected Components in an Undirected Graph"

     1 public class Solution {
     2     int[][] moves = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
     3 
     4     public List<Integer> numIslands2(int m, int n, int[][] positions) {
     5         List<Integer> res = new ArrayList<Integer>();
     6         if (positions.length==0) return res;
     7 
     8         int[][] grid = new int[m][n];
     9         List<Integer> rootList = new ArrayList<Integer>();
           // This is IMPORTANT: we need make sure root 1 is on index 1.
    10 rootList.add(0); 11 int index = 1; 12 int count = 0; 13 14 for (int i=0;i<positions.length;i++){ 15 int[] p = positions[i]; 16 for (int j=0;j<4;j++){ 17 int[] next = new int[]{p[0]+moves[j][0],p[1]+moves[j][1]}; 18 if (next[0]>=0 && next[0]<m && next[1]>=0 && next[1]<n && grid[next[0]][next[1]]!=0){ 19 // if p has not assigned index 20 if (grid[p[0]][p[1]]==0){ 21 grid[p[0]][p[1]] = grid[next[0]][next[1]]; 22 } else { 23 // merge root indexs of p and next. 24 int rootP = findRoot(rootList,grid[p[0]][p[1]]); 25 int rootNext = findRoot(rootList,grid[next[0]][next[1]]); 26 if (rootP!=rootNext){ 27 rootList.set(rootNext,rootP); 28 count--; 29 } 30 } 31 } 32 } 33 // if no merge happens, assign a new root index to p 34 if (grid[p[0]][p[1]]==0){ 35 rootList.add(index); 36 grid[p[0]][p[1]] = index++; 37 count++; 38 } 39 40 res.add(count); 41 } 42 43 return res; 44 } 45 46 public int findRoot(List<Integer> rootList, int id){ 47 while (rootList.get(id)!=id){ 48 int root = rootList.get(id); 49 rootList.set(id,rootList.get(root)); 50 id = rootList.get(id); 51 } 52 return id; 53 } 54 55 56 }
  • 相关阅读:
    [CareerCup] 4.6 Find Next Node in a BST 寻找二叉搜索树中下一个节点
    Android 接入支付宝支付实现
    Android 设置软键盘搜索键以及监听搜索键点击事件
    Android 应用监听自身卸载,弹出用户反馈调查
    ndk制作so库,ndk-build不是内部或外部命令。。。的错误
    Error: Your project contains C++ files but it is not using a supported native build system
    Android开发之——依赖冲突Program type already present
    基于Python的开源人脸识别库:离线识别率高达99.38%
    Android5.0以后,materialDesign风格的加阴影和裁剪效果
    Android 5.0 以上监听网络变化
  • 原文地址:https://www.cnblogs.com/lishiblog/p/5772630.html
Copyright © 2011-2022 走看看