Remove Duplicates from Sorted Array II

最后更新于:2022-04-02 01:08:33

# Remove Duplicates from Sorted Array II ### Source - leetcode: [Remove Duplicates from Sorted Array II | LeetCode OJ](https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/) - lintcode: [(101) Remove Duplicates from Sorted Array II](http://www.lintcode.com/en/problem/remove-duplicates-from-sorted-array-ii/) ~~~ Follow up for "Remove Duplicates": What if duplicates are allowed at most twice? For example, Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3]. Example ~~~ ### 题解 在上题基础上加了限制条件元素最多可重复出现两次。~~因此可以在原题的基础上添加一变量跟踪元素重复出现的次数,小于指定值时执行赋值操作。但是需要注意的是重复出现次数`occurence`的初始值(从1开始,而不是0)和reset的时机。~~这种方法比较复杂,谢谢 @meishenme 提供的简洁方法,核心思想仍然是两根指针,只不过此时新索引自增的条件是当前遍历的数组值和『新索引』或者『新索引-1』两者之一不同。 ### C++ ~~~ class Solution { public: /** * @param A: a list of integers * @return : return an integer */ int removeDuplicates(vector &nums) { if (nums.size() <= 2) return nums.size(); int len = nums.size(); int newIndex = 1; for (int i = 2; i < len; ++i) { if (nums[i] != nums[newIndex] || nums[i] != nums[newIndex - 1]) { ++newIndex; nums[newIndex] = nums[i]; } } return newIndex + 1; } }; ~~~ ### Java ~~~ public class Solution { /** * @param A: a array of integers * @return : return an integer */ public int removeDuplicates(int[] nums) { if (nums == null) return -1; if (nums.length <= 2) return nums.length; int newIndex = 1; for (int i = 2; i < nums.length; i++) { if (nums[i] != nums[newIndex] || nums[i] != nums[newIndex - 1]) { newIndex++; nums[newIndex] = nums[i]; } } return newIndex + 1; } } ~~~ ### 源码分析 遍历数组时 i 从2开始,newIndex 初始化为1便于分析。 ### 复杂度分析 时间复杂度 O(n)O(n)O(n), 空间复杂度 O(1)O(1)O(1).
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