19.2.15 [LeetCode 80] 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:

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]);
}

题意

去除重复元素使得数组中同一个元素最多出现两次

题解

最简单的想法当然就是直接使用erase，然而不够快

 1 class Solution {
 2 public:
 3     int removeDuplicates(vector<int>& nums) {
 4         int i = 0;
 5         while (i < nums.size()) {
 6             if (i >= 2 && nums[i] == nums[i - 1] && nums[i] == nums[i - 2])
 7                 nums.erase(nums.begin() + i);
 8             else i++;
 9         }
10         return nums.size();
11     }
12 };

View Code

后面用了种快慢指针的方式（早该想到的……）

 1 class Solution {
 2 public:
 3     int removeDuplicates(vector<int>& nums) {
 4         int n = nums.size();
 5         if (n <= 2)return n;
 6         int fast = 1, slow = 0, cnt = 0;
 7         while (fast < n) {
 8             if (fast == 0 || fast >= 1 && nums[fast] != nums[fast - 1]) {
 9                 cnt = 0;
10                 nums[++slow] = nums[fast++];
11             }
12             else if (cnt == 0) {
13                 cnt++;
14                 nums[++slow] = nums[fast++];
15             }
16             else
17                 fast++;
18         }
19         return slow+1;
20     }
21 };

View Code

19.2.15 [LeetCode 80] Remove Duplicates from Sorted Array II

原文：https://www.cnblogs.com/yalphait/p/10385004.html