LeetCode:Remove Elementとvector:erase()

LeetCode:Remove Elementとvectorのerase()
1、タイトル:Given an array and a value,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 in place with constant memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.
Example: Given input array nums = [3,2,2,3], val = 3
Your function should return length = 2, with the first two elements of nums being 2.
Hint:
Try two pointers. Did you use the property of “the order of elements can be changed”? What happens when the elements to remove are rare?
2、コード:

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        sort(nums.begin(),nums.end());// val     
        for(auto iter=nums.begin();iter!=nums.end();++iter)
        {
            if(*iter==val)
            {
                auto iter1=iter+1;//      
                for(;iter1!=nums.end()&&*iter1==val;++iter1);
                nums.erase(iter,iter1);//[b,e) 
                break;
            }
        }
        return nums.size();
    }
};

3、総括:まず並べ替えて、もともとテーマは順序を保つことを要求しないで、それから[b,e)を確定して、eraseを便利にします.注意:この関数は要素を削除した後の次の要素の反復器を返します.