Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

 

双指针。O(n)

i标定已处理范围，[0, i]下标的内容都是已经unique且有序的。最后返回的size即i+1。

j遍历指针，在找到一个与i当前指着的对象不同的对象时，换到i后面的位置，让独特串增长。

遍历逻辑：指的对象相同时，只加j。指的对象不同时，加i，赋值，加j。 

 

类似题目：Move Zeroes  https://www.cnblogs.com/jasminemzy/p/9479214.html

 

实现：

class Solution {    public int removeDuplicates(int[] nums) {        if (nums == null || nums.length == 0) {            return 0;        }        int i = 0, j = 0;        while (i < nums.length && j < nums.length) {            if (nums[i] == nums[j]) {                j++;            } else {                nums[++i] = nums[j++];            }        }        return i + 1;    }}