80. Remove Duplicates from Sorted Array II

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本文介绍了一种在原地修改输入数组的方法，使得数组中每个元素最多出现两次，并返回新的长度。通过实例展示了如何将数组[1,1,1,2,2,3]处理为[1,1,2,2,3]，并讨论了不使用额外空间的解决方案。

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https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/description/

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]);
}

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) <= 1:
            return len(nums)
        count = 1
        curr = nums[0]
        repeat = False
        i = 1
        while i < len(nums):
            if nums[i] == curr and not repeat:
                repeat = True
                count += 1
                i += 1
            elif nums[i] != curr:
                repeat = False
                curr = nums[i]
                count += 1
                i += 1
            else:
                nums.pop(i)
        return count