문제
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).
내 풀이
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class Solution {
public void rotate(int[] nums, int k) {
// 풀이법: 배열 크기보다 회전 수가 많다면 배열 크기보다 같거나 작도록 수정한다. (한 바퀴 돌아서 반영되지 않기에)
// 회전된 배열에서 인덱스가 원래 배열에서 어떤 인덱스인지 역으로 계산한다.
// if added index is larger than array's size, adjust it to fit it.
// example, [0,1,2,3,4], k=3 -> [2,3,4,0,1] 2's index : 2->5 (rotate k times)-> 5-5(array's length) =0
int[] rotated = new int[nums.length];
//if K is larger than array's length, make it smaller or equal to array's length
if(nums.length < k){
while(nums.length <k){
k -= nums.length;
}
}
// rotated index -> find original index
// 회전 후 인덱스 = 원래 인덱스 + 회전수 / (원래 인덱스 +회전수가 배열 크기보다 크다면) 원래 인덱스 + 회전수 - 배열 길이
// 역산하여 회전 후 배열에 들어갈 값의 원래 배열의 인덱스를 알아낸다.
int index =0;
for(int i=0; i<nums.length; i++){
index= (i-k) <0 ? i-k+nums.length : i-k;
rotated[i] = nums[index];
}
// copy rotated array to original array
// 원래 배열을 재정렬 해야하기에 회전 후 값 저장한 배열을 복사한다.
for(int i=0; i<nums.length; i++){
nums[i] = rotated[i];
}
}
}
}
}