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[Hyeri1ee] WEEK 01 solutions #2379
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,14 @@ | ||
| import java.util.*; | ||
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| class Solution { | ||
| Set<Integer> save = new HashSet<>();//O(1)에 이미 등장했는지 판별 위함 | ||
| public boolean containsDuplicate(int[] nums) { | ||
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| for(int i=0; i<nums.length; i++){ | ||
| if (!save.contains(nums[i])) save.add(nums[i]); | ||
| else return true; | ||
| } | ||
| return false; | ||
| } | ||
| } | ||
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,24 @@ | ||
| import java.util.*; | ||
| //goal : 두개의 인덱스 반환 | ||
| class Solution { | ||
| static int[] answer = new int[2]; | ||
| public int[] twoSum(int[] nums, int target) { | ||
| Map<Integer, Integer> maps = new HashMap<>();//num , index | ||
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. HashMap을 사용해서 {값 -> 인덱스}를 저장해 두고 한 번의 순회로 정답을 찾는 방식이라 시간 복잡도가 O(n)으로 잘 설계된 풀이로 보입니다! |
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| for(int i = 0; i < nums.length; i++){ | ||
| int t = target - nums[i]; //새로운 타겟 | ||
| //O(1)으로 다음 부분 찾기 | ||
| if (maps.containsKey(t)){ | ||
| answer[0] = i; | ||
| answer[1] = maps.get(t); | ||
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| Arrays.sort(answer); | ||
| return answer; | ||
| } | ||
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| maps.put(nums[i], i); | ||
| } | ||
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| return answer; | ||
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| } | ||
| } | ||
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여기서는
Set.add의 반환값을 활용해서처럼 한 줄로 줄여볼 수도 있을 것 같습니다!
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set의 add를 통해 한 줄로 줄여볼 수도 있네요! 조언 감사합니다