[array] leetCode-27. Remove Element - Easy

27. Remove Element – Easy

descrition

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 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

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

解析

核心思想:双指针;注意需求的分析,根据需求进一步优化设计。

方法 1

参见 code: int removeElementKeepOrder(vector

快慢指针,icur 指向新数组的结尾,i 遍历数组,只要当前值不等于 val 则进行复制。

方法 2

注意到题目的两个前提条件:(1)数组中元素的位置可以改变;(2)超过最新长度 length 后面的元素无关紧要。

这两个条件使得算法进一步优化成为可能。比如当 nums=[1,2,3,5,4],val=4,如果使用方法 1,将产生 4 次赋值操作。而 nums=[4,1,2,3,5],val=4 时,也将产生 4 次赋值操作,实际上我们可以直接将 4 直接删掉的。

还是两个指针,只不过这是两个对立的指针,icur 从前往后,指向当前要检查的数,在此之前的数都是满足要求的,当 nums[icur] 不满足要求时,值需要将 nums[iend] 赋值给它,并将 iend 自减即可, 这相当于之间删除 val。虽然时间复杂度还是 O(n),但实际上赋值操作的次数只等于需要删除的元素个数,当需要删除的元素个数很少时,算法很高效。

code


#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution{
public:
    int removeElement(vector<int>& nums, int val){
        //return removeElementKeepOrder(nums, val);
        return removeElementChangePlace(nums, val);
    }

    // time-O(n), space-O(1)
    int removeElementKeepOrder(vector<int>& nums, int val){
        int icur = 0;
        for(int i=0; i<nums.size(); i++){
            if(nums[i] != val){
                nums[icur++] = nums[i];
            }
        }

        return icur; // return new length
    }

    // time-O(n). In this approach, the number of assignment operation is 
    // equal to the number of elements to remove. So it is more efficent if 
    // elements to remove are rare.
    // Note: this optimization is based on the condition of 
    //     (1) The order of elements can be changed
    //     (2) It doesn't matter what you leave beyond the new length.   
    // space-O(1),
    int removeElementChangePlace(vector<int>& nums, int val){
        int icur = 0;
        int iend = nums.size() - 1;
        while(icur<=iend){
            if(nums[icur] == val){
                // move nums[icur] to the end and delete which operation is equal to subtract iend.
                // note: icur can't be decrease, because the current element
                // dosen't check 
                nums[icur] = nums[iend];
                iend--;
            }else{
                icur++;
            }
        }

        return iend+1;
    }
};

int main()
{
    return 0;
}

作者:fanling999 链接:https://www.cnblogs.com/fanling999/

    原文作者:fanling999
    原文地址: https://www.cnblogs.com/fanling999/p/7829482.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞

发表评论

电子邮件地址不会被公开。 必填项已用*标注