拓扑排序

要看数据结构的话，给一个大佬的博客，很详细

http://www.cnblogs.com/skywang12345/p/3603935.html

正题

我们现在来侃一下拓扑排序，简而言之，就是从有向无环图中找出要给有序的线性序列

把没有结点指向的结点先拿出来，根据先后排序，再把它从邻接矩阵中删除，递归调用

就这样，一直进行下去，就会得到最终结果。

给出我的代码

package com.java.demo.算法.图.拓扑排序.test;


import org.junit.Test;

import java.util.ArrayList;
import java.util.List;

public class Graph {
    private char[] points;
    private int edges[][];
    private int num;    //结点的个数

    public Graph() {
    }

    public Graph(char[] points, char[][] edges) {
        this.points = points;
        this.num = points.length;
        this.edges = new int[num][num];
        for (char[] edge : edges) {
            this.edges[getPosition(edge[0])][getPosition(edge[1])] = 1;
        }
    }

    //打印邻接矩阵信息
    public void print(){
        System.out.print("  ");
        for (char point : points) {
            System.out.print(point+" ");
        }
        System.out.println();
        for (int i=0;i<num;i++){
            System.out.print(points[i]+" ");
            for (int j=0;j<num;j++){
                System.out.print(edges[i][j]+" ");
            }
            System.out.println();
        }
    }

    public int getPosition(char ch){
        for (int i=0;i<num;i++){
            if (ch == points[i]){
                return i;
            }
        }
        return -1;
    }


    public void doFilter(int edges[][],char points[],List<Character> rets){
        int num = points.length;    //存放结点的数量
        List<Integer> list = new ArrayList<>();
        for (int j=0;j<num;j++){
            for (int i=0;i<num;i++){
                if (edges[i][j] == 1){
                    list.add(j);
                    break;
                }
            }
        }
        for (int i=0;i<num;i++){
            if (!list.contains(i)){
                rets.add(points[i]);
            }
        }
        num = list.size();
        if (num == 0){
            return;
        }
        int newEdges[][] = new int[num][num];
        char newPoints[] = new char[num];
        for (int i=0;i<num;i++){
            newPoints[i] = points[list.get(i)];
        }
        for (int i=0;i<num;i++){
            for (int j=0;j<num;j++){
                newEdges[i][j] = edges[list.get(i)][list.get(j)];
            }
        }
        doFilter(newEdges,newPoints,rets);
    }


    //拓扑排序
    public void topSort(){
        List<Character> rets = new ArrayList<>();
        doFilter(edges,points,rets);

        for (Character ret : rets) {
            System.out.print(ret+" ");
        }

//        for (int i=0;i<num;i++){
//            for (int j=0;j<num;j++){
//                if (edges[j][i] == 1){
//                    queue.add(points[i]);
//                    break;
//                }
//            }
//        }
//
//        for (Character character : queue) {
//            System.out.println();
//            System.out.print(character+" ");
//        }
    }

    public static void main(String[] args) {
        char points[] = {'A','B','C','D','E'};
        char edges[][] = {
                {'A','B'},
                {'B','D'},
                {'C','D'},
                {'D','E'},
        };
        Graph t = new Graph(points, edges);

//        t.print();
////
        t.topSort();
    }
}