Leetcode之Remove Duplicates from Sorted List I II

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本文介绍如何在不使用额外空间的情况下，从有序数组中移除重复元素，并保持每个元素只出现一次或两次的方法。通过使用快慢指针技巧，实现原地修改数组并返回新的有效长度。

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26. Remove Duplicates from Sorted Array

Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.
Example 1:
Given nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn’t matter what you leave beyond the returned length.

解题思路

使用快慢指针来记录遍历的坐标，最开始时两个指针都指向第一个数字，如果两个指针指的数字相同，则快指针向前走一步，如果不同，则两个指针都向前走一步，这样当快指针走完整个数组后，慢指针当前的坐标加1就是数组中不同数字的个数。同时，题目还要求以O(1)的空间复杂度，原地修改原数组，使得没有重复的值。

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int j = 0, n = nums.size(); //j为慢指针，i为快指针
        for (int i = 0; i < n; ++i) {
            if (nums[i] != nums[j]) nums[++j] = nums[i];
        }
        return j + 1;
    }
};

80. Remove Duplicates from Sorted Array II

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Example 1:
Given 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 respectively.
It doesn’t matter what you leave beyond the returned length.

解题思路

这道题是之前那道 Remove Duplicates from Sorted Array 有序数组中去除重复项的延续，这里允许最多重复的次数是两次，那么我们就需要用一个变量count来记录还允许有几次重复，count初始化为1，如果出现过一次重复，则count递减1，那么下次再出现重复，快指针直接前进一步，如果这时候不是重复的，则count恢复1，由于整个数组是有序的，所以一旦出现不重复的数，则一定比这个数大，此数之后不会再有重复项。

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size()<=2)
            return nums.size();
        int pre=0,cur=1,count=1;
        while(cur<nums.size()){
            if(nums[pre]==nums[cur] && count == 0) //下次再出现重复，快指针直接前进一步
                ++cur;
            else{
                if(nums[pre]==nums[cur]) //如果出现过一次重复，则count递减1
                    --count;
                else
                    count=1;  //如果这时候不是重复的，则count恢复1
                nums[++pre]=nums[cur++];
            }
        }
        return pre+1;
    }
};