Remove Duplicates from Sorted Array II
描述
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]
分析
加一个变量记录一下元素出现的次数即可。这题因为是已经排序的数组，所以一个变量即可解决。如果是没有排序的数组，则需要引入一个hashmap来记录出现次数。
代码1
// Remove Duplicates from Sorted Array II
// Time complexity: O(n), Space Complexity: O(1)
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.size() <= 2) return nums.size();
int index = 2;
for (int i = 2; i < nums.size(); i++){
if (nums[i] != nums[index - 2])
nums[index++] = nums[i];
}
return index;
}
};
代码2
下面是一个更简洁的版本。上面的代码略长，不过扩展性好一些，例如将occur < 2改为occur < 3，就变成了允许重复最多3次。
// Remove Duplicates from Sorted Array II
// Time Complexity: O(n), Space Complexity: O(1)
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
const int n = nums.size();
int index = 0;
for (int i = 0; i < n; ++i) {
if (i > 0 && i < n - 1 && nums[i] == nums[i - 1] && nums[i] == nums[i + 1])
continue;
nums[index++] = nums[i];
}
return index;
}
};