（双指针）leetcode 26、 27、203、283

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最新推荐文章于 2022-11-15 11:26:55 发布

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分类专栏： leetcode 文章标签： leetcode 数组 twopointer 指针

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本文介绍了几种常见的数组与链表操作算法，包括去除重复元素、移除特定值及零元素整理等。通过实例展示了如何利用双指针技术高效地进行这些操作，同时保持内存占用最小。

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26 Remove Duplicates from Sorted Array

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

使用双指针思想。
class Solution {

public:
    int removeDuplicates(vector<int>& nums) {
        int n=nums.size();
        if(n<2) return n;
        int count=1;
        for(int i=1;i<n;i++)
        {
            if(nums[i]!=nums[i-1]) nums[count++]=nums[i];
        }
        return count;
    }
};

27. Remove Element

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.

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

203. Remove Linked List Elements

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        if(head==NULL) return head;
        ListNode* pHead=head;
        ListNode* countNode=new ListNode(0);
        countNode->next=head;
        while(pHead!=NULL)
        {
            if(pHead->val!=val) 
            {
                countNode=countNode->next;
                countNode->val=pHead->val;
             }
            pHead=pHead->next;
        }
        countNode->next=NULL;
        if(head->val==val) return NULL;//注意特殊情况，当元素全部是所给值的时候
        return head;
    }
};

283. Move Zeroes

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:

You must do this in-place without making a copy of the array.
Minimize the total number of operations.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int n=nums.size();
        if(n==0) return ;
        int count=0;
        for(int i=0;i<n;i++)
        {
            if(nums[i]!=0) nums[count++]=nums[i];
        }
        while(count<n)
        nums[count++]=0;
        
    }
};