Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,1,2,2,3],

Your function should return length = [code]5

, with the first five elements of

nums

being

1, 1, 2, 2

and 3 respectively. It doesn't matter what you leave beyond the returned length.[/code]

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length = [code]7

, with the first seven elements of

nums

being modified to 

0

, 0, 1, 1, 2, 3 and 3 respectively. It doesn't matter what values are set beyond the returned length. [/code]

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

26. Remove Duplicates from Sorted Array 的拓展，如果允许最多两次重复，返回去除之后的长度。

解法：同样使用双指针，增加一个变量记录出现的字符数量。

Java:

public class Solution {
public int removeDuplicates(int[] A) {
if (A == null || A.length == 0)
return 0;

int pre = A[0];
boolean flag = false;
int count = 0;

// index for updating
int o = 1;

for (int i = 1; i < A.length; i++) {
int curr = A[i];

if (curr == pre) {
if (!flag) {
flag = true;
A[o++] = curr;

continue;
} else {
count++;
}
} else {
pre = curr;
A[o++] = curr;
flag = false;
}
}

return A.length - count;
}
}

Java:

public class Solution {
public int removeDuplicates(int[] A) {
if (A.length <= 2)
return A.length;

int prev = 1; // point to previous
int curr = 2; // point to current

while (curr < A.length) {
if (A[curr] == A[prev] && A[curr] == A[prev - 1]) {
curr++;
} else {
prev++;
A[prev] = A[curr];
curr++;
}
}

return prev + 1;
}
}　　

Python:

class Solution:
# @param a list of integers
# @return an integer
def removeDuplicates(self, A):
if not A:
return 0

last, i, same = 0, 1, False
while i < len(A):
if A[last] != A[i] or not same:
same = A[last] == A[i]
last += 1
A[last] = A[i]
i += 1

return last + 1　　

C++：

class Solution {
public:
int removeDuplicates(int A[], int n) {
if (n <= 2) return n;
int pre = 0, cur = 1, count = 1;
while (cur < n) {
if (A[pre] == A[cur] && count == 0) ++cur;
else {
if (A[pre] == A[cur]) --count;
else count = 1;
A[++pre] = A[cur++];
}
}
return pre + 1;
}
};

　　

类似题目：

[LeetCode] 26. Remove Duplicates from Sorted Array 有序数组中去除重复项

 

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