lintcode521- Remove Duplicate Numbers in Array- easy

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

Given nums = [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.

1.O(n) time: 用HashSet来确认有过没，有过就swap(切记这里要做一个 i--的操作，因为你把后面没有检验过的数换到这里了，要重新确认当前位置)；没有就加入set，继续。

2.O(n) time:用HashSet。一开始就把所有数字都加入HashSet。之后遍历set（切记set没有set.get(i)方法，只能用iterator的那个冒号方法），放到数组的最前面。

3.O(nlogn)time, but no extra space: 先sort，把一样的数字聚集到一起。再用一个int cntDistct 来同时做计数和指针，如果找到独特的数，就直接替换到前面指针的位置。

1. HashSet+swap

public class Solution {
    /*
     * @param nums: an array of integers
     * @return: the number of unique integers
     */
    public int deduplication(int[] nums) {
        // write your code here
        Set<Integer> set = new HashSet<Integer>();
        int countDup = 0;
        for (int i = 0; i < nums.length - countDup; i++) {
            if (!set.contains(nums[i])) {
                set.add(nums[i]);
            } else {
                swap(nums, i, nums.length - 1 - countDup);
                countDup++;
                i--;
            }
        }
        return nums.length - countDup;
    }
    
    private void swap (int[] nums, int idx1, int idx2) {
        int temp = nums[idx1];
        nums[idx1] = nums[idx2];
        nums[idx2] = temp;
    }
}

2. HashSet

public class Solution {
    /*
     * @param nums: an array of integers
     * @return: the number of unique integers
     */
    public int deduplication(int[] nums) {
        // write your code here
        
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            set.add(nums[i]);
        }
        
        int i = 0;
        for (Integer number : set) {
            nums[i++] = number;
        }
        return set.size();
    }
}

3. sort

public class Solution {
    /*
     * @param nums: an array of integers
     * @return: the number of unique integers
     */
    public int deduplication(int[] nums) {
        // write your code here
        
        Arrays.sort(nums);
        int countDis = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            nums[countDis++] = nums[i];
        }
        return countDis;
    }
}