LeetCode-Easy-Remove Element

###原题目
```cpp
Given an array nums and a value val, remove all instances of that value in-place 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.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

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

It doesn't matter what you leave beyond the returned length.
Example 2:

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

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

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 = removeElement(nums, val);

// 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]);
}
```
###自己拿到手第一想法
其实已经想到现在做出来的这个想法的，但是想更简单一些，所以一直在想用一些比较好的条件判断，但是绕了很久都没有出来，所以现在还是用原来的标记数的想法来。
现在想法是先用sort排序，在经过for循环与val值进行对比，将首迭代器存入，然后标记依次加一，而后用erase删除即可
###网上优秀的代码
用了distance以及remove，将其中与value相同直接删除，再次用distance算出距离。
std::distance(nums.begin(),remove(nums.begin(),nums.end(),val));
remove容器操作的源代码
```cpp
template <class ForwardIterator, class T>
  ForwardIterator remove (ForwardIterator first, ForwardIterator last, const T& val)
{
  ForwardIterator result = first;
  while (first!=last) {
    if (!(*first == val)) {
      *result = move(*first);
      ++result;
    }
    ++first;
  }
  return result;
}
```
###自己的想法
还是基础不太扎实，必须不断的练习，然后不断地熟悉各种泛型算法，以及他们背后的实现原理。