[LeetCode#82]Remove Duplicates from Sorted Array II

最新推荐文章于 2025-08-13 23:36:12 发布

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原文链接：http://www.cnblogs.com/airwindow/p/4752253.html

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本文探讨了如何在已排序的数组中，处理最多允许出现两次重复元素的情况，通过分析错误的解决方案并提供正确的实现方式，旨在解决数组去重问题，确保每个元素在数组中最多出现两次。

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Problem:

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.

Analysis:

A wrong solution: 
<When you update on a array and check on the array, you must be careful about if you get the original data or updated date>
public int removeDuplicates(int[] nums) {
    if (nums == null || nums.length == 0)
        return 0;
    if (nums.length <= 2)
        return nums.length;
    int count = 2;
    for (int i = 2; i < nums.length; i++) {
        if (nums[i] == nums[i-1] && nums[i-1] == nums[i-2])
            continue;
        count++;
        nums[count-1] = nums[i];
    }
    return count;
}

Problem 1:
This solution is ugly!!! The code 
if (nums[i] == nums[i-1] && nums[i-1] == nums[i-2])
    continue;
count++;
nums[count-1] = nums[i];

The above code could be written into:
if !(nums[i] == nums[i-1] && nums[i-1] == nums[i-2])
    nums[count] = nums[i];
    count++;
    
Problem 2:
The solution has implemention logic error.
<When you update on a array and check on the same array, you must be careful about if you get the original data or updated date>
Cases:
1, 1, 1, 2, 2
After interation: i == 3,
1, 1, (2), 2, *2
At interation: i == 4
We could see 
nums[4] == nums[3] && nums[3] == nums[2]
Which is wrong!!! we replaced nums[2] with 2, but nums[3] still in it's original position. We lose the information of original nums[2].

How could we solve this problem??? 
A great idea: check if (nums[i] != nums[count-2])
Note: the count pointer always point to the next avaiable position.
nums[count-1] means the last element we place into nums.
nums[count-2] means the last two element we place into nums.

Keep on thing in mind, if the current element num[i] has already been appeared more than two times, it must be nums[count-1] and nums[count-2]. !!! And if nums[count-2] == nums[count], it means nums[count-2] must equal to nums[count-1]. 
If not, we could not skip it!
if (nums[i] != nums[count-2]) {
    nums[count] = nums[i];
    count++;
}

Genius thinking!

Solution:

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        if (nums.length <= 2)
            return nums.length;
        int count = 2;
        for (int i = 2; i < nums.length; i++) {
            if (nums[i] != nums[count-2]) {
                nums[count] = nums[i];
                count++;
            }
        }
        return count;
    }
}