26. Remove Duplicates from Sorted Array
Given a sorted array nums , remove the duplicates in- place such that each element appears only once and returns the new length.
Do not allocate extra space for another array, you must do this by modifying the input arrayin-place with O(1) extra memory.
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 a 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]);
}
Example 1:
Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: 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:
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 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.
Constraints:
0 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
nums
is sorted in ascending order.
# @lc code=start
using LeetCode
function remove_duplicates1!(nums::Vector{Int})::Int
j = 2
for i in 2: length(nums)
if nums[i] != nums[j - 1]
nums[j] = nums[i]
j += 1
end
end
return j - 1
end
# @lc code=end
remove_duplicates1! (generic function with 1 method)
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