线性表-26.Remove Duplicates from Sorted Array

最新推荐文章于 2018-08-12 14:27:16 发布

原创最新推荐文章于 2018-08-12 14:27:16 发布 · 511 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一种在不使用额外空间的情况下对已排序数组进行去重的方法，并提供了两种不同去重需求的具体实现：一种是每个元素只允许出现一次，另一种则是允许最多出现两次。

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 arrayA = [1,1,2],
Your function should return length = 2, and A is now[1,2].

题意:给定一个有序数列,确保每个元素只能出现一次,返回修改后的数组长度

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

类的题:80.Remove Duplicates from Sorted Araay II

Follow up for ”Remove Duplicates”: What if duplicates are allowed at most twice?
For example, Given sorted arrayA = [1,1,1,2,2,3],
Your function should return length = 5, and A is now[1,1,2,2,3]

题意:对数组进行去重操作,同一个元素只能最多出现两次,返回新数组大小,同时更新数组.就是将后面的数据覆盖前面的数据.

public class Solution {
    public int removeDuplicates(int[] nums) {
        int len = nums.length;
        if(len < 3){
            return len;
        }
        
        int index = 2;
        for(int i = 2; i<len; i++){
            if(nums[index-2] != nums[i])
                nums[index++] = nums[i];
        }
        return index;
        }
    }