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36 changes: 35 additions & 1 deletion leetcode3/이진희/3940_Limit Occurrences in Sorted Array.java
Original file line number Diff line number Diff line change
@@ -1,4 +1,38 @@
/*
/* 2차 풀이

1. 아이디어 : 정렬된 배열을 순회하며, 현재 값 개수가 k 이상이면 continue, 이하면 붙인다 (이전 풀이와 동일)

2. 시간복잡도 : O(N)

3. 자료구조/알고리즘 : 완전탐색

*/

class Solution {
public int[] limitOccurrences(int[] nums, int k) {
int cnt = 1;
int prev = nums[0];
int len = 0;
int[] ans = new int[nums.length];

ans[len++] = nums[0];

for(int i=1; i<nums.length; i++) {
if(prev != nums[i]) {
cnt = 1;
prev = nums[i];
}
else cnt++;

if(cnt>k) continue;
ans[len++] = nums[i];
}

return Arrays.copyOfRange(ans, 0, len);
}
}

/* 1차 풀이

1. 아이디어 : 배열을 순회하며, 현재 값 개수가 k 이상이면 continue, 이하면 붙인다

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