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 ofnumscontaining0,1,3,0, 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.
题目所要求的是要使用inplace方法，也就是只能在原值/原数组上更改，不能再开辟新的地址空间。
用C++很好的是vector是一个动态的，可以直接用erase（）函数把没用的删掉就行了，如果用C的话相对就会困难一些了。
class Solution {
public:
int removeElement(vector<int>& nums, int val) {
//使用in place操作是对这个值/数组本身进行操作
//https://blog.csdn.net/u011489043/article/details/70197020
for(int i=0;i<nums.size();i++){
if(nums[i]==val){
nums.erase(nums.begin()+i);
i--;
}
}
return nums.size();
}
};
注意，如果要删掉那个元素的话，需要把i回退。
vector的内部实现：https://blog.csdn.net/u012658346/article/details/50725933