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problem： 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].
受上题目的启发，如果n为零，则返回零，如果不为零，则num和i从1开始，当A[i] != A[i-1]时，我们就把A[i]赋值给A[num]，同时num++，最后num就是新数组的长度。
class Solution { public: int removeDuplicates(int A[], int n) { if (n == 0) return 0; int num = 1; for ( int i = 1; i < n; i++) { if(A[i - 1] != A[i]) A[num++] = A[i]; } return num; } };
2015/03/29:
Python:
class Solution: # @param a list of integers # @return an integer def removeDuplicates(self, A): if len(A) == 0: return 0 start = 0 for i in range(1, len(A)): if A[i] != A[start]: start += 1 A[start] = A[i] return start + 1
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