521.Remove Duplicate Numbers in Array

1.Description(Medium)

Given an array of integers, remove the duplicate numbers in it.

You should: 1. Do it in place in the array. 2. Move the unique numbers to the front of the array. 3. Return the total number of the unique numbers.

Notice

You don't need to keep the original order of the integers.

Example

Givennums=[1,3,1,4,4,2], you should:

1. Move duplicate integers to the tail of

   nums=>nums=[1,3,4,2,?,?].

2. Return the number of unique integers in

   nums=>4.

Actually we don't care about what you place in?, we only care about the part which has no duplicate integers.

Challenge

1. Do it in O(n) time complexity.

2. Do it in O(nlogn) time without extra space.

2.Code

Solution 1：HashSet ，O(n) time, O(n) space

//Solution 1: O(n) time, O(n) space
    public int deduplication(int[] nums) {
        //HashMap<Integer,Boolean> map=new HashMap<Integer,Boolean>();
        HashSet<Integer> set=new HashSet<Integer>();
        for(int i=0;i<nums.length;i++){
            if(!set.contains(nums[i])){
                set.add(nums[i]);
            }
        }
        int result=0;
        for(Integer element:set){
            nums[result]=element;
            result++;
        }
        return result;
    }

Solution 2：In place O(nlogn) time, O(1) extra space

http://www.jiuzhang.com/solutions/remove-duplicate-numbers-in-array/

public int deduplication(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        Arrays.sort(nums);
        int len = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != nums[len]) {
                nums[++len] = nums[i];
            }
        }
        return len + 1;
    }

这个题目分array有没有sorted.

1.如果已经是sorted的话，最好的方法是就是用一个prev记录前面的，然后遍历，直接加进result,不必要浪费O(n) 的space 去建立一个hashset.

2.如果没有sorted,就是这个题，第一种是时间花费O(n)，空间花费O(n)建一个hashset.

                                                  第二种是