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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