leetcode-Remove Elements

最新推荐文章于 2022-04-11 13:43:40 发布

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最新推荐文章于 2022-04-11 13:43:40 发布

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分类专栏： leetcode系列文章标签：算法 leetcode java

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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

一次遍历，将非零元素往前提即可。

public class Solution {
    public void moveZeroes(int[] nums) {
     int j = 0;
     for(int i = 0; i < nums.length; i++) {
        if(nums[i] != 0) {
            nums[j++] = nums[i];
        }
    }
     while(j<nums.length){
            nums[j++]=0;
    }
    }
}

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.

同理，一次遍历，对于非val元素，往前提。因为返回前n个非val 元素的数组，后面的不需要设置。

public class Solution {
    public int removeElement(int[] nums, int val) {
        int index = 0;
        for (int i=0 ; i<nums.length ; i++){
            if (nums[i] != val){
                nums[index] = nums[i];
                index++;
                
            }
        }
        return index;
    }
}

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.

因为是已排好序的数组，所以，只要判断该数与最后记录的num[index] 是否相同。

public class Solution {
    public int removeDuplicates(int[] nums) {
        int index =0 ;
        for (int i=1; i <nums.length; i++){
            if (nums[i] != nums[index]){
                index++;
                nums[index] = nums[i];
            }  
        }
        return index+1;
    }
}

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

可以通过递归解决，只需考虑当前节点的值是否与所要删除的值相同。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null ) return null;
        /**if (head.val == val ){
            head=removeElements(head.next,val);
        }else{
            head.next = removeElements(head.next,val);
        }
         return head;  
        */ 
         
        if (head == null) return null;
        head.next = removeElements(head.next, val);
        return head.val == val ? head.next : head;
        
    }
}