LeetCode26 Remove Duplicates from Sorted Array

问题描述：
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]);
}
题源：here；完整代码：here
思路：
构造两个指针，第一个记录出现的唯一数据个数，也可以理解为需要返回的长度；第二个依次遍历输入数据，如果与第一个指针指向的数不同，就将第一个指针下移一位，并置为第二个指针指向的数。因为输入数组已经排序，所以只需要遍历一遍输入就可以了。下面有两个实现版本。
代码1

    int removeDuplicates_1(vector<int>& nums) {
        if (!nums.size()) return 0;
        int len = 0, loc = 0;
        for (int i = 0; i < nums.size() - 1; i++){
            if (nums[i] == nums[i + 1]) continue;
            nums[len] = nums[loc];
            len++; loc = i + 1;
        }
        nums[len] = nums[loc];
        return len + 1;
    }

代码2

    int removeDuplicates_2(vector<int>& nums) {
        if (!nums.size()) return 0;
        int len = 0;
        for (int i = 0; i < nums.size(); i++){
            if (nums[len] == nums[i]) continue;
            else nums[++len] = nums[i];
        }
        return len + 1;
    }

吐槽：越是简单的题是不是题干越长。。。