Cracking coding interview(2.1)去除LinkedList中的重复元素

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2.1 Write code to remove duplicates from an unsorted linked list.
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How would you solve this problem if a temporary buffer is not allowed?

import java.util.LinkedList;import java.util.Iterator;import java.util.Collections;import java.util.Hashtable;public class Solution{//brute-force time complexity:O(n^2) space complexity:O(1)public static void removeDuplicate1(LinkedList<Integer> list){for(int i=0;i < list.size()-1;i++)for(int j=i+1;j < list.size();)if(list.get(i) == list.get(j))list.remove(j);elsej++;}//could't keep order time complexity:O(nlogn) space complexity:O(1)public static void removeDuplicate2(LinkedList<Integer> list){//sortCollections.sort(list);if(list.size() >= 2){for(int i=0;i < list.size()-1;){if(list.get(i) == list.get(i+1))list.remove(i+1);elsei++;}}}//1.keep order 2.time complexity:O(n) 3.space complexity:O(n): (worst case)public static void removeDuplicate3(LinkedList<Integer> list){Hashtable<Integer, String> hash = new Hashtable<Integer, String>();//lookup hashtable to delete repeat elementsfor(int i=0;i < list.size();){if(hash.containsKey(list.get(i)))list.remove(i);else{hash.put(list.get(i), "");i++;}}}private static void printLinkedList(LinkedList<Integer> list){Iterator it = list.iterator();while(it.hasNext()){System.out.print((Integer)it.next()+" ");}System.out.println();}public static void main(String[] args){LinkedList<Integer> list = new LinkedList<Integer>();list.add(6);list.add(2);list.add(2);list.add(3);list.add(1);list.add(4);list.add(2);list.add(3);list.add(7);list.add(2);list.add(2);list.add(10);Solution.printLinkedList(list);Solution.removeDuplicate3(list);Solution.printLinkedList(list);}}

1.brute-force time complexity: O(n^2) space complxity:O(1), 输出元素保持原有顺序

2.sort：time complexity：O(nlogn) space complexity:O(1), 输出元素为排序后结果

3.hashtable：time complexity：O(n) space complexity:O(n), 输出元素保持原有顺序

类似问题:http://blog.csdn.net/u011559205/article/details/38125405

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