LeetCode-Remove Duplicates from Sorted Array II

最新推荐文章于 2022-04-06 16:22:53 发布

转载最新推荐文章于 2022-04-06 16:22:53 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/incrediblechangshuo/p/5931642.html

本文介绍了一种算法，该算法可以在允许最多两个重复项的情况下移除数组中的多余重复元素，并返回处理后的有效长度。通过两个不同的实现方式展示了如何就地修改输入数组。

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.

感觉这道题的oa有点问题。。。先这样吧

public class Solution {
    public int removeDuplicates(int[] nums) {
        
        int res=0;
        int len=nums.length;
        int set=0;
        
        for(int i=0; i<len; i++){
            if(i==0){
                set++;
                res++;
            }
            else{
                if(nums[i]==nums[i-1]){
                    set++;
                    if(set<=2){
                        res++;
                    }
                }
                else{
                    set=1;
                    res++;
                }
            }

        }
        return res;
    }
}

要in place 改变原数组：

public class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums==null || nums.length==0){
            return 0;
        }
        int startPosition=0;
        boolean isRepeated=false;
        for(int i=1; i<nums.length; i++){
            if(nums[i] != nums[startPosition]){
                isRepeated=false;
                startPosition++;
                nums[startPosition] = nums[i];
            }
            else{
                if(isRepeated == false){
                    startPosition++;
                    nums[startPosition]=nums[i];
                    isRepeated=true;
                }
            }
        }
        return startPosition+1;
        
    }
}

转载于:https://www.cnblogs.com/incrediblechangshuo/p/5931642.html