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));
    }
}