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Leetcode总结之Graph

时间:2016-04-03 01:49:52      阅读:251      评论:0      收藏:0      [点我收藏+]

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package Graph;

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;

public class GraphProblem {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

    }
    /*
     * 207. Course Schedule
     * 2016-3-28 by jyuan
     * BFS:典型的拓扑排序。原理也很简单,在一个有向图中,每次找到一个没有前驱节点的节点(也就是入度为0的节点),
     * 然后把它指向其他节点的边都去掉,重复这个过程(BFS),直到所有节点已被找到,或者没有符合条件的节点(如果图中有环存在)。。
     * DFS的解法,也需要建立有向图,还是用二维数组来建立,和BFS不同的是,我们像现在需要一个一维数组visit来记录访问状态,
     * 大体思路是,先建立好有向图,然后从第一个门课开始,找其可构成哪门课,暂时将当前课程标记为已访问,
     * 然后对新得到的课程调用DFS递归,直到出现新的课程已经访问过了,则返回false,没有冲突的话返回true,然后把标记为已访问的课程改为未访问
     * http://www.jyuan92.com/blog/leetcode-course-schedule/
     */
    public boolean canFinishWithBFS(int numCourses, int[][] prerequisites) {
        if (null == prerequisites || numCourses == 0 || prerequisites.length == 0) {
            return true;
        }
        int[] preCourses = new int[numCourses];
        // store the in-degree #
        for (int[] prerequisite : prerequisites) {
            preCourses[prerequisite[0]]++;
        }
        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < preCourses.length; i++) {
            if (preCourses[i] == 0) {
                queue.add(i);
            }
        }
        int numOfPreCourse = 0;
        while (!queue.isEmpty()) {
            int top = queue.poll();
            numOfPreCourse++;
            for (int[] prerequisite : prerequisites) {
                if (prerequisite[1] == top) {//Find all the neighbor
                    preCourses[prerequisite[0]]--;
                    if (preCourses[prerequisite[0]] == 0) {
                        queue.add(prerequisite[0]);
                    }
                }
            }
        }
        return numOfPreCourse == numCourses;
    }
    /*
     * 210 Course ScheduleII
     * 2016-3-28 by jyuan
     * http://www.jyuan92.com/blog/leetcode-course-schedule-ii/
     */
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] res = new int[numCourses];//区别1,多了一个装order的array
        int[] preCourses = new int[numCourses];
        // store the in-degree #
        for (int[] prerequisite : prerequisites) {
            preCourses[prerequisite[0]]++;
        }
        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < preCourses.length; i++) {
            if (preCourses[i] == 0) {
                queue.add(i);
            }
        }
        int numOfPreCourse = 0;
        int i = 0;//区别2,多了一个i来遍历res
        while (!queue.isEmpty()) {
            int top = queue.poll();
            res[i++] = top;
            numOfPreCourse++;
            for (int[] prerequisite : prerequisites) {
                if (prerequisite[1] == top) {
                    preCourses[prerequisite[0]]--;
                    if (preCourses[prerequisite[0]] == 0) {
                        queue.add(prerequisite[0]);
                    }
                }
            }
        }//最后再多一个判断,就是返回是否是需要return新的
        if (numOfPreCourse == numCourses) {
            return res;
        } else {
            return new int[0];
        }
    }
    /*
     * 下面给出207的DFS方法:
     * DFS的思路就简单一点,就是首先构造图的Map,每一个点包含一个list,表明这个点作为这个list
     * 每一门课程的前备课程以待后用。那么主体思路就是每一个点为起点做一次dfs,知道底端,
     * 如果没有问题就continue,如果有问题直接return false。
     * 那么在主体部分,几个状态,如果为-1表示已经访问过肯定return false,如果为1表示这条路
     * 已经是通路,已经验证完毕,所有连接这个点的直接return true。
     */
    public boolean canFinishWithDFS(int numCourses, int[][] prerequisites) {
        if (null == prerequisites || numCourses == 0 || prerequisites.length == 0) {
            return true;
        }
        int[] isVisited = new int[numCourses];
        Map<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
        for (int[] prerequisite : prerequisites) {
            if (!map.containsKey(prerequisite[1])) {
                map.put(prerequisite[1], new ArrayList<Integer>());
            }
            map.get(prerequisite[1]).add(prerequisite[0]);
        }
        for (int i = 0; i < numCourses; i++) {
            if (!dfs(isVisited, map, i)) {
                return false;
            }
        }
        return true;
    }
    private boolean dfs(int[] isVisited, Map<Integer, ArrayList<Integer>> map, int index) {
        if (isVisited[index] == -1) {
            return false;
        }
        if (isVisited[index] == 1) {
            return true;
        }
        isVisited[index] = -1;//假如最后一个点进来,先赋值-1,然后没有map的值,所以直接为1,以后所有遇到这个点的都是通路
        if (map.containsKey(index)) {
            for (int value : map.get(index)) {
                if (!dfs(isVisited, map, value)) {
                    return false;
                }
            }
        }
        isVisited[index] = 1;
        return true;
    }
    /*
     * 下面给出210的DFS方法:
     */
    int i;//区别1全局变量i
    public int[] findOrderWithDFS(int numCourses, int[][] prerequisites) {
        i = numCourses - 1;
        int[] res = new int[numCourses];
        int[] isVisited = new int[numCourses];
        Map<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
        for (int[] prerequisite : prerequisites) {
            if (!map.containsKey(prerequisite[1])) {
                map.put(prerequisite[1], new ArrayList<Integer>());
            }
            map.get(prerequisite[1]).add(prerequisite[0]);
        }
        for (int i = 0; i < numCourses; i++) {
            if (!dfs(map, isVisited, res, i)) {
                return new int[0];
            }
        }
        return res;
    }     
    private boolean dfs(Map<Integer, ArrayList<Integer>> map, int[] isVisited, int[] res, int index) {
        if (isVisited[index] == -1) {
            return false;
        }
        if (isVisited[index] == 1) {
            return true;
        }
        isVisited[index] = -1;
        if (map.containsKey(index)) {
            for (int value : map.get(index)) {
                if (!dfs(map, isVisited, res, value)) {
                    return false;
                }
            }
        }
        res[i--] = index;
        isVisited[index] = 1;
        return true;
    }
    /*
     * 261.Graph Valid Tree
     * 2016-3-29 by Buttercola
     * 和上面的BFS方法一样,不过这里是无向图,所以需要两个点都要存对方
     * 图要形成树必须有两个方面,一个是没有cycle另一个是没有孤立的点
     * Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), 
     * write a function to check whether these edges make up a valid tree.
     * Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.
     * Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.
     */
    private boolean valid(int n, int[][] edges) {
        // build the graph using adjacent list
        List<Set<Integer>> graph = new ArrayList<Set<Integer>>();   
        for(int i = 0; i < n; i++)
            graph.add(new HashSet<Integer>());
        for(int[] edge : edges) {
            graph.get(edge[0]).add(edge[1]);
            graph.get(edge[1]).add(edge[0]);
        }
        // no cycle
        boolean[] visited = new boolean[n];
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.add(0);
        while(!queue.isEmpty()) {
            int node = queue.poll();
            if(visited[node])
                return false;
            visited[node] = true;
            for(int neighbor : graph.get(node)) {
                queue.offer(neighbor);
                graph.get(neighbor).remove((Integer)node);
            }
        }
        // fully connected
        for(boolean result : visited){
            if(!result)
                return false;
        }
        return true;
    }
    /*
     * 310. Minimum Height Trees
     * 2016-3-29 by Mingyang
     * 同样的方法构建无向图
     * 这里运用的一个层层剥丝的方式,一层一层的来,去掉不要的
     */
     public List<Integer> findMinHeightTrees(int n, int[][] edges) {
            if (n == 1) return Collections.singletonList(0);
            List<Set<Integer>> adj = new ArrayList<Set<Integer>>(n);
            for (int i = 0; i < n; ++i) adj.add(new HashSet<Integer>());
            for (int[] edge : edges) {
                adj.get(edge[0]).add(edge[1]);
                adj.get(edge[1]).add(edge[0]);
            }
            List<Integer> leaves = new ArrayList<Integer>();
            for (int i = 0; i < n; ++i)
                if (adj.get(i).size() == 1) leaves.add(i);
            while (n > 2) {
                n -= leaves.size();
                List<Integer> newLeaves = new ArrayList<Integer>();
                for (int i : leaves) {
                    int j = adj.get(i).iterator().next();
                    adj.get(j).remove(i);
                    if (adj.get(j).size() == 1) newLeaves.add(j);
                }
                leaves = newLeaves;
            }
            return leaves;
        }
    /*
     * 332. Reconstruct Itinerary 
     * 2016-3-14 by Mingyang
     * 这道题给我们一堆飞机票,让我们建立一个行程单,如果有多种方法,取其中字母顺序小的那种方法。
     * 这道题的本质是有向图的遍历问题,那么LeetCode关于有向图的题只有两道Course Schedule和Course Schedule II,
     * 而那两道是关于有向图的顶点的遍历的,而本题是关于有向图的边的遍历。
     * 每张机票都是有向图的一条边,我们需要找出一条经过所有边的路径,那么DFS不是我们的不二选择。
     * 代码前半段都在准备工作。把所有的tickets全部装进一个map,string对应了一个priority queue
     * 后半段dfs中,找出第一个顺序的结果。最终输出是从最后一个开始,一个一个往前退的情况下addFirst加起来的
     */
    HashMap<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>();
    LinkedList<String> result = new LinkedList<String>();
    public List<String> findItinerary(String[][] tickets) {
        for (String[] ticket : tickets) {
            if (!map.containsKey(ticket[0])) {
                PriorityQueue<String> q = new PriorityQueue<String>();
                map.put(ticket[0], q);
            }
            map.get(ticket[0]).offer(ticket[1]);
        }
        dfs("JFK");
        return result;
    }
    public void dfs(String s) {
        PriorityQueue<String> q = map.get(s);
        while (q != null && !q.isEmpty()) {
            dfs(q.poll());
        }
        result.addFirst(s);
    }
    /*
     * 323    Number of Connected Components in an Undirected Graph
     * 2016-4-2 by Mingyang
     * Given n nodes labeled from 0 to n - 1 and
     *  a list of undirected edges (each edge is a pair of nodes), 
     *  write a function to find the number of connected components in an undirected graph.
     *  You can assume that no duplicate edges will appear in edges. 
     *  Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
     */
     public int countComponents(int n, int[][] edges) {
            if (n <= 1) {
                return n;
            }
            List<List<Integer>> adjList = new ArrayList<List<Integer>>();
            for (int i = 0; i < n; i++) {
                adjList.add(new ArrayList<Integer>());
            }
            for (int[] edge : edges) {
                adjList.get(edge[0]).add(edge[1]);
                adjList.get(edge[1]).add(edge[0]);
            }
            boolean[] visited = new boolean[n];
            int count = 0;
            for (int i = 0; i < n; i++) {
                if (!visited[i]) {
                    count++;
                    dfs(visited, i, adjList);
                }
            }
            return count;
        }
        public void dfs(boolean[] visited, int index, List<List<Integer>> adjList) {
            visited[index] = true;
            for (int i : adjList.get(index)) {
                if (!visited[i]) {
                    dfs(visited, i, adjList);
                }
            }
        }
}

 

Leetcode总结之Graph

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原文地址:http://www.cnblogs.com/zmyvszk/p/5348786.html

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