LeetCode80. 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:

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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]);
}

Two pointers

1

my code in cpp

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int removeDuplicates(vector<int>& nums) {
if (nums.empty()) return 0;
int _new = 0, _old =0, count = 1;
for (; ++_old < nums.size();)
{
if (nums[_old] != nums[_new]) count = 1;
else count++;
if (count <= 2) nums[++_new] = nums[_old];
}
return ++_new;}

总结

  1. 双指针,一个遍历原来的,一个更新新数组。并且计数,只更新2以内的数字,如果遇到不重复的将计数设为1,否则计数加一。
  2. 入口检查空值。返回的是长度不是索引。
Thanks for Support.