[Lintcode] 172. Remove Element

172. Remove Element

Description

Given an array and a value, remove all occurrences of that value in place and return the new length.

The order of elements can be changed, and the elements after the new length don’t matter.

Example
Given an array [0,4,4,0,0,2,4,4], value=4

return 4 and front four elements of the array is [0,0,0,2]

我的代码：

class Solution:
    """
    @param: A: A list of integers
    @param: elem: An integer
    @return: The new length after remove
    """
    def removeElement(self, A, elem):
        # write your code here
        f = 0
        while f != len(A):
            if A[f] == elem:
                A.pop(f)
            else:
                f += 1
        return f

方法二：

class Solution:
    def removeElement(self, A, elem):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """        
        l = len(A)
        head = 0
        tail = len(A) - 1
        if tail == 0:
            if A[0] == elem:
                return 0
            else:
                return 1
        while head < tail:
            while A[head] != elem:
                head += 1
                if head==len(A):
                    return len(A)
            while A[tail] == elem:
                tail -= 1
                if tail<0:
                    break
            if head > tail:
                break
            tmp = A[head]
            A[head] = A[tail]
            A[tail] = tmp
        return head

思路：

时间复杂度： O(n)
一个很麻烦的问题，因为题目讲得没那么清楚。虽然只需要返回更新后的长度，但是A本身也需要实时更新。除了pop以外，其实也可以调换顺序.

我遇到的一个问题，涉及到了python中赋值，拷贝到底是如何实现的。https://my.oschina.net/leejun2005/blog/145911