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[LeetCode] Remove Element

时间:2017-07-07 12:30:17      阅读:267      评论:0      收藏:0      [点我收藏+]

题目内容

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.

题目思路

题目难度:easy

此题虽然不难,但是蕴含的思想非常重要。这个题目就是数组快慢指针的思想来源。设定两个指针,i和ii,其中i是慢指针,ii是快指针。初始化的时候,ii和i均等于0。快指针遍历所有的数组,当nums[ii]!=val,则将nums[ii]赋值给nums[i],然后i+=1。此时i就是删除了指定元素后的数组长度。

Python代码

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

 

[LeetCode] Remove Element

原文:http://www.cnblogs.com/chengyuanqi/p/7131308.html

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