LeetCode 刷题记录27. Remove Element

题目：
27. Remove Element

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]);
}
解法1：
利用stl 中的erase函数 注意该函数删除元素后，返回值指向已删除元素的下一个位置
或者remove函数
c++:

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
         for(vector<int> :: iterator it = nums.begin(); it != nums.end(); ){
            if(*it == val){
                nums.erase(it);
            } else {
                it++;
            }
        }
        return nums.size();
    }
};

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        return distance(nums.begin(), remove(nums.begin(), nums.end(), val));
    }
};

解法2：
用index变量计数 ，遍历数组，一旦遇到不是目标元素值，覆盖原数组，index+1
c++:

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int n = nums.size();
        int index = 0;
        for(int i = 0; i < n; ++i){
            if(nums[i] != val){
                nums[index++] = nums[i];
            }
        }
        return index;
    }
};

java:

class Solution {
    public int removeElement(int[] nums, int val) {
        int n = nums.length;
        int index = 0;
        for(int i = 0; i < n; ++i){
            if(nums[i] != val){
                nums[index++] = nums[i];
            }
        }
        return index;
    }
}

python:
Python中可以直接遍历元素，不用用索引值

class Solution(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        n = len(nums)
        index = 0
        for num in nums:
            if num != val:
                nums[index] = num
                index += 1
            
        
        return index

直接用pop或者remove函数 但是复杂度不是O(N)
python里面pop，remove和del 三者的用法区别

class Solution(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        while val in nums:
            nums.pop(nums.index(val))
           # nums.remove(val)
            
        
        return len(nums)