LeetCode刷题【Array】 Remove Duplicates from Sorted Array II

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本文探讨了在数组中允许元素重复出现最多两次的情况下如何去除多余重复项的问题，并提供了两种高效实现方法。

题目：

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.

解决方法一： Runtime: 1 ms

public class Solution {
    public int removeDuplicates(int[] nums) {
			int count=1;
			int i =0;
			int index =0;
			int sum=0;
			while(i<=nums.length-1){
				if(i-1>=0){
					if(nums[i]==nums[i-1]){
						count++;
						if(count<=2){
							nums[index+1]=nums[i];
							index++;
						}
					}else if(nums[i]!=nums[i-1]){
						nums[index+1]=nums[i];
						index++;
						sum+=count>2?2:count;
						count=1;
					}
				}
				i++;
			}			
			return sum+(count>2?2:count);
    }
}

解决方法二： Runtime: 1 ms

public int removeDuplicates(int[] nums) {
    int i = 0;
    for (int n : nums)
        if (i < 2 || n > nums[i-2])
            nums[i++] = n;
    return i;
}

参考：

【1】https://leetcode.com/