Leetcode 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.

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

双指针

和leetcode 26的解法类似, 构造扫描指针j和插入指针i, 当j扫描到满足条件的值时就插入到i中.

本题需要两个变量存储信息. currentNumber储存当前的值, 来判断是否和前面重复. currentCount储存currentNumber重复的次数. \

首先判断当前的nums[j]和currentNumber是否相等. 若相等则判断重复次数, 若重复的次数超过2次, 则j直接自增. 不超过2次, 则currentCount自增, 并将nums[j++]插入到nums[i++].

若不相等, 则更新currentNumber为nums[j], 并将nums[j++]插入到nums[i++]

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size() <= 2)
            return nums.size();
        int i = 1;
        int j = 1;
        int currentNumber = nums[0];
        int currentCount = 1;
        while(j < nums.size())
        {
            if(nums[j] != currentNumber)
            {
                currentNumber = nums[j];
                currentCount = 1;
                nums[i++] = nums[j++];
            }
            else
            {
                ++currentCount;
                if(currentCount == 2)
                {
                    nums[i++] = nums[j++];
                }
                else if(currentCount > 2)
                {
                    ++j;
                }
            }
        }
        return i;
    }
};