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dohyeon2
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Feb 28, 2026
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간단한 문제지만, Rust를 재미있게 잘 살펴볼 수 있었습니다.
좋은 풀이 감사합니다.
| for (i, &num) in nums.iter().enumerate() { | ||
| let complement = target - num; | ||
| if let Some(&j) = indices.get(&complement) { | ||
| return vec![j as i32, i as i32]; |
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'이 부분에서 왜 정렬을 안하지?'라고 생각했는데
You can return the answer in any order.
최면에 걸린 것 처럼 정렬을 해버렸는데,,
이런 조건이 있네요.
심지어 순차적으로 조회하며 index를 반환하는거라
정렬을 안해도 결과가 같다는 당연한 사실을 알게되었습니다..
| // TC: O(n) | ||
| // SC: O(n) | ||
| impl Solution { | ||
| pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> { |
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Vec<i32>의 의미는 알겠는데,
i32라는 표현이 꽤나 무섭네요.
확실히 다른 인기있는 언어들에 비해 저수준을 다룬다는 걸 알 수 있네요.
| pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> { | ||
| let mut indices = HashMap::new(); | ||
| for (i, &num) in nums.iter().enumerate() { | ||
| let complement = target - num; |
| let mut indices = HashMap::new(); | ||
| for (i, &num) in nums.iter().enumerate() { | ||
| let complement = target - num; | ||
| if let Some(&j) = indices.get(&complement) { |
dohyeon2
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Feb 28, 2026
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