删除有序数组中的重复元素 Remove Duplicates from Sorted Array

题目源自于Leetcode。

题目：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].

思路：左/右两个指针控制着原地的新/旧数组。

注意时刻关注着右指针是否到达末尾。

class Solution {
public:
int removeDuplicates(int A[], int n) {
if(n<=1)
return n;
int left, right;
left = 0;
right = 1;
while(right<n)
{
while(right<n && A[right] == A[left])
right++;
if(right == n)
break;

left++;
A[left] = A[right];
right++;
}
return left+1;
}
};

或者用for循环也一样。

class Solution {
public:
int removeDuplicates(int A[], int n) {
if(n == 0 || A == NULL)
return 0;

int index = 0;
for(int i=1;i<n;i++)
{
if(A[index] != A[i])
{
++index;
A[index] = A[i];
}
}
return index+1;
}
};

如果想偷懒使用STL算法的话，可以先unique()将重复元素仍在后面，然后再返回非重复元素的个数。但是这样的实现效率并没有直接双指针快。

class Solution {
public:
int removeDuplicates(int A[], int n) {
return distance(A, unique(A, A+n));
}
};