Given a sorted array, 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 in place with constant memory.
For example,
Given input array 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 new length.
Solution:
這道題的技巧是數(shù)組已經(jīng)排序了，所有的duplicated的元素都是相鄰的。因此可以使用two pointers的思路窗怒，用一個i指針遍歷所有元素的同時，用另一個j指針指向當前被拿來作比較的元素，如果i指向的元素等于j指向的元素，則什么都不做吮铭；如果i指向的元素不等于j指向的元素，意味著從此以后i指向的元素永遠不會再與j指向的相等颅停。因此可以將i指向的元素放在j指向元素的后一個位置并向后移動j指針（被拿來作比較的元素變成原來的后一個元素了）谓晌。
public class Solution
{
public int removeDuplicates(int[] nums)
{
int j = 0;
for(int i = 0; i < nums.length; i ++)
{
if(nums[j] != nums[i])
{
nums[j + 1] = nums[i];
j ++;
}
}
return j + 1;
}
}
