标签：应该 访问 rem return 代码 做了 design ase argument
题目原文：
Given a set of n integers S = {0,1,…,N-1}and a sequence of requests of the following form:
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
1 import edu.princeton.cs.algs4.StdOut; 2 3 public class Successor { 4 private int num; 5 private int[] id; 6 private boolean[] isRemove; 7 8 public Successor(int n){ 9 num = n; 10 id = new int[n]; 11 isRemove = new boolean[n]; 12 for (int i = 0; i < n; i++) { 13 id[i] = i; 14 isRemove[i] = false; 15 } 16 } 17 18 public int find(int p) { 19 while (p != id[p]) 20 p = id[p]; 21 return p; 22 } 23 24 public void union(int p, int q) { 25 //此处的union取较大根 26 int pRoot = find(p); 27 int qRoot = find(q); 28 if (pRoot == qRoot) 29 return; 30 else if (pRoot < qRoot) 31 id[pRoot] = qRoot; 32 else 33 id[qRoot] = pRoot; 34 } 35 36 public void remove(int x) { 37 isRemove[x] = true; 38 //判断相邻节点是否也被remove掉了，如果remove掉就union 39 if (x>0 && isRemove[x-1]){ 40 union(x,x-1); 41 } 42 if (x<num-1 && isRemove[x+1]){ 43 union(x,x+1); 44 } 45 } 46 47 public int getSuccessor(int x) { 48 if(x<0 || x>num-1){//越界异常 49 throw new IllegalArgumentException("访问越界!"); 50 }else if(isRemove[x]){ 51 if(find(x)+1 > num-1) //x以及大于x的数都被remove掉了，返回-1 52 return -1; 53 else //所有remove数集中最大值+1，就是successor 54 return find(x)+1; 55 }else {//x未被remove，就返回x自身 56 return x; 57 } 58 } 59 60 public static void main(String[] args) { 61 Successor successor = new Successor(10); 62 successor.remove(2); 63 successor.remove(4); 64 successor.remove(3); 65 StdOut.println("the successor is : " + successor.getSuccessor(3)); 66 successor.remove(7); 67 successor.remove(9); 68 StdOut.println("the successor is : " + successor.getSuccessor(9)); 69 } 70 }
Coursera Algorithms week1 练习测验3：Successor with delete
标签：应该 访问 rem return 代码 做了 design ase argument
原文地址：http://www.cnblogs.com/evasean/p/7204566.html
