题目：
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array 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. It doesn't matter what you leave beyond the new length.
链接：
https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/#/description
5/31/2017
这道题做了好多遍才对。不应该
1 public class Solution { 2 public int removeDuplicates(int[] nums) { 3 if (nums == null || nums.length == 0) return 0; 4 int copyIndex = 1; 5 int duplicateCount = 0; 6 for (int i = 1; i < nums.length; i++) { 7 if (nums[i - 1] != nums[i]) { 8 nums[copyIndex] = nums[i]; 9 copyIndex++; 10 duplicateCount = 0; 11 } else { 12 duplicateCount++; 13 if (duplicateCount == 1) { 14 nums[copyIndex] = nums[i]; 15 copyIndex++; 16 } 17 } 18 } 19 return copyIndex; 20 } 21 }
别人的答案
https://discuss.leetcode.com/topic/17180/3-6-easy-lines-c-java-python-ruby
1 public int removeDuplicates(int[] nums) { 2 int i = 0; 3 for (int n : nums) 4 if (i < 2 || n > nums[i-2]) 5 nums[i++] = n; 6 return i; 7 }
最多k次的一般解
1 int removeDuplicates(int A[], int n, int k) { 2 3 if (n <= k) return n; 4 5 int i = 1, j = 1; 6 int cnt = 1; 7 while (j < n) { 8 if (A[j] != A[j-1]) { 9 cnt = 1; 10 A[i++] = A[j]; 11 } 12 else { 13 if (cnt < k) { 14 A[i++] = A[j]; 15 cnt++; 16 } 17 } 18 ++j; 19 } 20 return i; 21 }
https://discuss.leetcode.com/topic/46519/short-and-simple-java-solution-easy-to-understand
更多讨论：
https://discuss.leetcode.com/category/88/remove-duplicates-from-sorted-array-ii