80. Remove Duplicates from Sorted Array II

problem 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]);
}

algorithm thought

这题目其实很简单，怎么样都能做得出。只是有些人做的简单有些人做的复杂。首先用for range语句，得到nums所有的数字。然后和nums[pos-2]相比，如果大于的话，那么当前肯定要加，但是如果等于的话，那肯定就是前面已经放置了两个相同的数，就可以不用再写了。

code

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int pos=0;
        for(auto num:nums){
            if(pos<2||num>nums[pos-2]){
                nums[pos++]=num;
            }
        }
        return pos;
    }
};

algorithm analysis

一次遍历出结果，时间复杂度是O(n)