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.
思路1:
相比于题目Remove Duplicates from Sorted Array,该题目允许两个重复值存在,所以设置一个标志位flag,当第二次插入该元素的时候令flag=true,下次如果flag为true,就不插入后面的元素了。
思路2:
除了借助位置指示i和j外,借助另一个位置指示k,来找到所有相同的元素,然后把相同元素第二次放入数组。
代码实现1:
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size() < 1) return 0;
int j = 0, i = 1;
bool flag = false;
while(i < nums.size())
{
if(nums[i] == nums[j] && !flag)
{
nums[++j] = nums[i];
flag = true;
}
else if(nums[i] != nums[j])
{
nums[++j] = nums[i];
flag = false;
}
++i;
}
return j+1;
}
};
代码实现2:
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size() < 1) return 0;
int j = 0, i = 1;
while(i < nums.size())
{
if(nums[i] == nums[j])
{
int k = i + 1;
while(k < nums.size() && nums[k] == nums[i])
++k;
nums[++j] = nums[i];
i = k;
}
else
nums[++j] = nums[i++];
}
return j+1;
}
};