# 26.remove-duplicates-from-sorted-array

## Statement

``````int[] nums = […]; // 输入数组
int[] expectedNums = […]; // 长度正确的期望答案
int k = removeDuplicates(nums); // 调用
assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
assert nums[i] == expectedNums[i];
}``````

``````输入：nums = [1,1,2]

``````

``````输入：nums = [0,0,1,1,1,2,2,3,3,4]

``````

• `0 <= nums.length <= 3 * 104`
• `-104 <= nums[i] <= 104`
• `nums` 已按 升序 排列

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.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array `nums`. More formally, if there are `k` elements after removing the duplicates, then the first `k` elements of `nums` should hold the final result. It does not matter what you leave beyond the first `k` elements.

Return `k` after placing the final result in the first `k` slots of `nums`.

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

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 * 104`
• `-100 <= nums[i] <= 100`
• `nums` is sorted in non-decreasing order.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
int removeDuplicates(vector<int> &nums) {
int ix = 1;

for (int i = 1; i < nums.size(); i++) {
if (nums[i] != nums[i - 1]) {
nums[ix++] = nums[i];
}
}

return ix;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````