LeetCode Remove Duplicates from Sorted Array 移除有序数列的重复元素

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

小心令人抓狂的下标！

方法一：

class Solution {public:    int removeDuplicates(int A[], int n) {        if(n<=1) return n;        int j = 0;        for (int i = 1; i < n; ++i) {            if (A[i] != A[j]) {                A[++j] = A[i];            }        }        return ++j;    }};

方法二：

class Solution {public:int removeDuplicates(int A[], int n) {if(n<=1) return n;int dup = 0;int i = 1;for (; i < n; i++){if(A[i] == A[i-1]) dup++;else if(dup != 0){for (int j = i;  j < n; j++){A[j-dup] = A[j];}n -= dup;i -= dup;dup = 0;}}if(dup != 0){A[n-dup-1] = A[n-1];n -= dup;dup = 0;}return n;}};

C++ STL

template <class ForwardIterator>  ForwardIterator unique (ForwardIterator first, ForwardIterator last){  if (first==last) return last;  ForwardIterator result = first;  while (++first != last)  {    if (!(*result == *first))  // or: if (!pred(*result,*first)) for version (2)      *(++result)=*first;  }  return ++result;}

//2014-1-25class Solution {public:int removeDuplicates(int A[], int n) {int i = 1, j = 1;for ( ; j < n; j++){if (A[j] != A[j-1]) A[i++] = A[j];}return n? i:0;}};