[LeetCode] 80. Remove Duplicates from Sorted Array II 删除有序数组中的重复项之二

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length
int k = removeDuplicates(nums); // Calls your implementation
assert k == expectedNums.length;

for (int i = 0; i < k; i++) {

assert nums[i] == expectedNums[i];

}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

1 <= nums.length <= 3 * 10^4
-10^4 <= nums[i] <= 10^4
nums is sorted in non-decreasing order.

这道题是之前那道 Remove Duplicates from Sorted Array 的拓展，这里允许最多重复的次数是两次，那么可以用一个变量 cnt 来记录还允许有几次重复，cnt 初始化为1，如果出现过一次重复，则 cnt 递减1，那么下次再出现重复，快指针直接前进一步，如果这时候不是重复的，则 cnt 恢复1，由于整个数组是有序的，所以一旦出现不重复的数，则一定比这个数大，此数之后不会再有重复项。理清了上面的思路，则代码很好写了：

解法一：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int pre = 0, cur = 1, cnt = 1, n = nums.size();
        while (cur < n) {
            if (nums[pre] == nums[cur] && cnt == 0) ++cur;
            else {
                if (nums[pre] == nums[cur]) --cnt;
                else cnt = 1;
                nums[++pre] = nums[cur++];
            }
        }
        return nums.empty() ? 0 : pre + 1;
    }
};

这里其实也可以用类似于 Remove Duplicates from Sorted Array 中的解法三的模版，由于这里最多允许两次重复，那么当前的数字 num 只要跟上上个覆盖位置的数字 nusm[i-2] 比较，若 num 较大，则绝不会出现第三个重复数字（前提是数组是有序的），这样的话根本不需要管 nums[i-1] 是否重复，只要将重复个数控制在2个以内就可以了，参见代码如下：

解法二：

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