Coursera Algorithms week1 查并集 练习测验：3 Successor with delete

题目原文：

Given a set of n integers S = {0,1,…,N-1}and a sequence of requests of the following form:

• Remove x from S
• Find the successor of x: the smallest y in S such thaty>=x

design a data type so that all operations(except construction) take logarithmic time or better in the worst case.

分析

题目的要求有一个0~n-1的顺序排列序列S，从S中移除任意x，然后调用getSuccessor(x)，方法将返回一个y，这个y是剩余还在S中满足y>=x的最小的数。举例说明S={0,1,2,3,4,5,6,7,8,9}时

remove 6，那么getSuccessor(6)=7

remove 5，那么getSuccessor(5)=7

remove 3，那么getSuccessor(3)=4

remove 4，那么getSuccessor(4)=7

remove 7，那么getSuccessor(7)=8, getSuccessor(3)=8

而对于没有remove的数x，getSuccessor(x)应该等于几呢？题目没有说，那么就认为等于自身好了，接着上面，getSuccessor(2)=2

根据上面的例子，可以看出，实际上是把所有remove的数做了union，root为子集中的最大值，那么getSuccessor(x)实际就是获取remove数中的最大值+1，根据这个思路，代码如下

 import edu.princeton.cs.algs4.StdOut;

 public class Successor {
     private int num;
     private int[] id;
     private boolean[] isRemove;

     public Successor(int n){
         num = n;
         id = new int[n];
         isRemove = new boolean[n];
         for (int i = 0; i < n; i++) {
             id[i] = i;
             isRemove[i] = false;
         }
     }

     public int find(int p) {
         while (p != id[p])
             p = id[p];
         return p;
     }

     public void union(int p, int q) {
         //此处的union取较大根
         int pRoot = find(p);
         int qRoot = find(q);
         if (pRoot == qRoot)
             return;
         else if (pRoot < qRoot)
             id[pRoot] = qRoot;
         else
             id[qRoot] = pRoot;
     }

     public void remove(int x) {
         isRemove[x] = true;
         //判断相邻节点是否也被remove掉了，如果remove掉就union
         if (x>0 && isRemove[x-1]){
             union(x,x-1);
         }
         if (x<num-1 && isRemove[x+1]){
             union(x,x+1);
         }
     }

     public int getSuccessor(int x) {
         if(x<0 || x>num-1){//越界异常
             throw new IllegalArgumentException("访问越界!");
         }else if(isRemove[x]){
             if(find(x)+1 > num-1) //x以及大于x的数都被remove掉了，返回-1
                 return -1;
             else //所有remove数集中最大值+1，就是successor
                 return find(x)+1;
         }else {//x未被remove，就返回x自身
             return x;
         }
     }

     public static void main(String[] args) {
         Successor successor = new Successor(10);
         successor.remove(2);
         successor.remove(4);
         successor.remove(3);
         StdOut.println("the successor is : " + successor.getSuccessor(3));
         successor.remove(7);
         successor.remove(9);
         StdOut.println("the successor is : " + successor.getSuccessor(9));
     }
 }