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]
之前的想法是再加一个计数的变量就行了
int removeDeplicates1(int A[], int n)
{
int index = 0, count = 1;
for (int i = 1; i < n; i++)
{
if (A[index] != A[i])
{
A[index++] = A[i];
count = 1;
}
else
{
if (count <= 2)
{
A[index] = A[i];
count++;
}
}
}
return index + 1;
}
后来看到了一个更简洁的做法，这个想法很巧妙啊：因为是排好序的，所以只需拿第三个和第一个比，不同的话保存即可。
int removeDeplicates1(int A[], int n)
{
int index = 2;
for (int i = 2; i < n; i++)
{
if (A[index-2] != A[i])
{
A[index++] = A[i];
}
}
return index;
}
leetcode 之Remove Duplicates from Sorted Array（二）
原文：http://www.cnblogs.com/573177885qq/p/5451089.html