# [LeetCode] 26. Remove Duplicates from Sorted Array ☆(从有序数组中删除重复项)

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

## 描述

Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn‘t matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn‘t matter what values are set beyond the returned length.

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]);
}

```给定数组: nums = [1,1,2],

## 代码

```class Solution {
public int removeDuplicates(int[] nums) {
if (nums == null || nums.length <= 0) {
return 0;
}
if (nums.length < 2) {
return 1;
}
int pre = 0, cur = 0, n = nums.length;
while (cur < n) {
if (nums[pre] == nums[cur])
++cur;
else
nums[++pre] = nums[cur++];
}
return pre + 1;
}
}```

```class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.empty()) return 0;
int j = 0, n = nums.size();
for (int i = 0; i < n; ++i) {
if (nums[i] != nums[j]) nums[++j] = nums[i];
}
return j + 1;
}
};```

## [LeetCode] Remove Duplicates from Sorted Array II 有序数组中去除重复项之二

Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3]. 这道题是之前那道Remove Duplicates from Sorted Array 有序数组中

## [LeetCode] 80. Remove Duplicates from Sorted Array II ☆☆☆(从有序数组中删除重复项之二)

https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/discuss/27976/3-6-easy-lines-C%2B%2B-Java-Python-Ruby 描述 Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1

## [LeetCode] Find All Duplicates in an Array 找出数组中所有重复项

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,3,2,7,

## leetCode 26. Remove Duplicates from Sorted Array 数组

26. Remove Duplicates from Sorted Array 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 mem

## LeetCode 26. Remove Duplicates from Sorted Array

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 nums

## leetCode 26.Remove Duplicates from Sorted Array(删除数组重复点) 解题思路和方法

Remove Duplicates from Sorted Array 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.