Tag: array, level: easy

Solution:

Using two pointers, use first pointer to iterate the whole array, second point to the position without duplicates. 

When the first pointer get to an non-duplicated element, assign it the the second pointer, then move second pointer a step further.

 After the first pointer finish the iteration, the second pointer is at the position of non-duplicates.

Time: O(n), space O(1)

https://github.com/Blankj/awesome-java-leetcode/blob/master/note/026/README.md