80. Remove Duplicates from Sorted Array II

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每日一题 2019 - 04 - 11

题目：

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:

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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:

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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:

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// 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]);
}

解法：

这个题让清除掉数组中重复的元素，但是这里重复的元素可以存在两次，思路比较直观，设置双指针，一个用来记录当前指针的位置，一个用来记录重复元素出现的次数，同时删除元素可以直接使用 vector 中的erase()函数，但是在这里一定要记得更改待遍历vector的大小，也即

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for(int i = 1 ; i < nums.size() ; i ++ )

否则可能会越界。

代码：

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class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size() == 0)
        {
            return 0 ;
        }
        int size = nums.size();
        int i = 0 ; 
        int j = 1 ; 
        int temp = nums[0] ;
        for(int i = 1 ; i < nums.size() ; i ++ )
        {
            if(temp == nums[i] )
            {
                if( j < 2 )
                {
                    j ++ ;   
                    continue ;
                }
                else
                {
                    nums.erase(nums.begin() + i);
                    i -- ;
                    continue ;
                }
            }
            else
            {
                j = 1 ; 
                temp = nums[i] ;
            }
        }
        return nums.size() ;
    }
};
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