# 26. Remove Duplicates from Sorted Array

Given an integer array `nums` sorted in non-decreasing order , remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in`nums`.

Consider the number of unique elements of `nums` to be `k`, to get accepted, you need to do the following things:

• Change the array `nums` such that the first `k` elements of `nums` contain the unique elements in the order they were present in `nums` initially. The remaining elements of `nums` are not important as well as the size of `nums`.
• Return `k`.

Custom Judge:

The judge will test your solution with the following code:

``````int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;

for (int i = 0; i < k; i++) {

assert nums[i] == expectedNums[i];

}
``````

If all assertions pass, then your solution will be accepted.

Example 1:

``````**Input:** nums = [1,1,2]
**Output:** 2, nums = [1,2,_]
**Explanation:** Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
``````

Example 2:

``````**Input:** nums = [0,0,1,1,1,2,2,3,3,4]
**Output:** 5, nums = [0,1,2,3,4,_,_,_,_,_]
**Explanation:** Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
``````

Constraints:

• `1 <= nums.length <= 3 * 10^4`
• `-100 <= nums[i] <= 100`
• `nums` is sorted in non-decreasing order.

``````class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int pre = 0, cur = 0, n = nums.size();
while (cur < n) {
if (nums[pre] == nums[cur]) ++cur;
else nums[++pre] = nums[cur++];
}
return nums.empty() ? 0 : (pre + 1);
}
};
``````

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

``````class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int i = 0;
for (int num : nums) {
if (i < 1 || num > nums[i - 1]) {
nums[i++] = num;
}
}
return i;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/26

Remove Duplicates from Sorted List

Remove Duplicates from Sorted List II

Remove Duplicates from Sorted Array II

Remove Element

Apply Operations to an Array

Sum of Distances

https://leetcode.com/problems/remove-duplicates-from-sorted-array/

https://leetcode.com/problems/remove-duplicates-from-sorted-array/discuss/11757/My-Solution-%3A-Time-O(n)-Space-O(1)

LeetCode All in One 题目讲解汇总(持续更新中…)

（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，快快加入吧～）

|

Venmo 打赏

—|—

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation