Published Dec 10, 2021
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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.
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.
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).
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).
- 0 <= nums.length <= 3 * 104
- -100 <= nums[i] <= 100
- nums is sorted in non-decreasing order.
function removeDuplicates(nums: number[]): number {
let curIndex = 0, resIndex = 0;
// iterate through the array
for(; curIndex < nums.length; curIndex++){
// move to 1 if it is 0, this also helps handle edge case
if(curIndex === 0) {
resIndex++
continue;
}
// if we find duplicates, ignore
if(nums[curIndex] === nums[curIndex - 1]){
continue;
} else {
// if we find unqiue, store the value at unique index to increase it by 1
nums[resIndex] = nums[curIndex];
resIndex++;
}
}
return resIndex;
};