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[DaleSeo] WEEK 10 Solutions #2588
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,33 @@ | ||
| // TC: O(log n) | ||
| // SC: O(1) | ||
| impl Solution { | ||
| pub fn search(nums: Vec<i32>, target: i32) -> i32 { | ||
| let mut low = 0i32; | ||
| let mut high = nums.len() as i32 - 1; | ||
|
|
||
| while low <= high { | ||
| let mid = low + (high - low) / 2; | ||
| let mid_val = nums[mid as usize]; | ||
|
|
||
| if mid_val == target { | ||
| return mid; | ||
| } | ||
|
|
||
| if nums[low as usize] <= mid_val { | ||
| if nums[low as usize] <= target && target < mid_val { | ||
| high = mid - 1; | ||
| } else { | ||
| low = mid + 1; | ||
| } | ||
| } else { | ||
| if mid_val < target && target <= nums[high as usize] { | ||
| low = mid + 1; | ||
| } else { | ||
| high = mid - 1; | ||
| } | ||
| } | ||
| } | ||
|
|
||
| -1 | ||
| } | ||
| } |
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🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 이진 탐색을 활용하여 배열의 회전 여부와 타겟 위치를 판단하며, 배열을 반복적으로 반으로 나누어 탐색 범위를 좁혀 나가는 방식이다. 배열의 크기를 반으로 줄이기 때문에 시간 복잡도는 로그에 비례한다. 추가적인 공간은 변수 몇 개만 사용하므로 상수 공간이다.
개선 제안: 현재 구현이 적절해 보입니다.