leetcode--Remove Duplicates from Sorted Array II-优快云博客

本文详细介绍了如何在已排序的数组中去除最多出现两次的重复元素，通过双指针技巧实现高效的元素去重。适用于前端开发、后端开发等领域的数组操作场景。

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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 ofnums being 1, 1, 2, 2 and3. It doesn't matter what you leave beyond the new length.

题意：移除排序数组中的重复元素，一个元素最多出现两次

分类：数组，双指针

解法1：和leetcode--Remove Duplicates from Sorted Array一样，使用双指针，但是每次覆盖时，判断是否由两个以上重复，如果是，覆盖两次，否则覆盖一次

public class Solution {
   public int removeDuplicates(int[] nums) {
        int cur = Integer.MIN_VALUE;
        int count = 0;
        int sum = 0;
        for(int i=0;i<nums.length;i++){
        	if(cur==nums[i]){
        		if(count==1){
        			nums[sum] = nums[i]; 
        			count++;
        			sum++;        			
        		}else{            			
        			count = 0;
        			cur = nums[i];        			
        		}
        	}else{
        		cur = nums[i];
        		nums[sum] = nums[i]; 
        		sum++;
        		count = 1;
        	}        	
        }
        return sum;
    }

}

public class Solution {
    public int removeDuplicates(int[] nums) {
        int count = 0;
        int len = nums.length;
        if(len==0) return 0;
        int pre = 0;
        int cur = 0;
        while(cur<len){
            if(nums[cur]==nums[pre]){
                cur++;
            }else{
                nums[count++] = nums[pre];
                if(cur-pre>=2){
                    nums[count++] = nums[pre];
                }
                pre = cur;
            }
        }
        nums[count++] = nums[cur-1];
        if(cur-pre>=2){
            nums[count++] = nums[cur-1];
        }
        return count;
    }
}