LeetCode-Remove Duplicates from Sorted Array

Description：
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.

Clarification:
Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

题意：有一个已排好序的数组，将其中的重复数字去除，要求空间复杂度为O(1)；

解法：由于数组是已排好序的，我们可以使用一个变量len来记录此时的不重复数组的长度和此时不重复数组的最后一个数的下标，当遍历数组遇到不重复数字时添加到此下标的后一个数的位置；

class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums.length == 0){
            return 0;
        }//空数组
        int len = 0;
        for(int i=1; i<nums.length; i++){
            if(nums[len] != nums[i]){
                len++;
                nums[len] = nums[i];//不重复的数组存储到前面
            }
        }
        return len+1;
    }
}