[Lintcode] 172. Remove Element

最新推荐文章于 2019-07-25 11:51:28 发布

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最新推荐文章于 2019-07-25 11:51:28 发布

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原文链接：http://www.cnblogs.com/siriusli/p/10353412.html

本文提供了LeetCode上第172题Remove Element的两种解决方案，详细讲解了如何在给定数组和特定值的情况下，移除所有该值的出现，并返回新的数组长度。第一种方法使用while循环和pop函数，第二种方法则采用双指针技巧，提高了效率。

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