1.Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
2.Do not allocate extra space for another array, you must do this in place with constant memory.
3.For example, Given input array A = [1,1,2],
Your function should return length = 2, and A is now [1,2].
分析：
大概题意是，给你一个已经排好序的数组，把数组中重复的数去掉，然后重新排列数组。
思路1：先去掉重复的数，然后排序。不过效率你懂的。
思路2：怎么也没有想到的做法，记住了，记住了！
//思路1：我想很多人也会跟我一样吧，嘿嘿。
class Solution{
public:
int removeDuplicates(int A[], int n){
int len = n;
int i,j;
for (i=0;i<n;i++){
j = i;
while(A[j] == A[++i]){//将相同的数置0
A[i]=0;
len--;
}
if (A[j] != A[i]){//i的指针往回移动一位，避免两次i++
i--;
}
}
//排序
for (i=1;i<len;i++){
if (A[i] == 0){
j= i+1;
while(A[j] == 0){
j++;
}
A[i] = A[j];
A[j++] = 0;
}
}
return len;
}
};
//思路2：大神的做法，总是这么让我心旷神怡
class Solution{
public:
int removeDuplicates2(int A[], int n){
if (n == 0 || n == 1)
{
return n;
}
int index = 0;
for(int i= 1;i<n;i++){
if (A[index] != A[i])
{
A[++index] = A[i];
}
}
return index + 1;
}
};