[6주차] 지상일 백준 문제 2002, 9252, 15486, 15565, 2616, 7453 풀이 #14
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| Map<String, Integer> map = new HashMap<>(); //차와 현재 위치를 저장할 맵 | ||
| for (int i = 1; i <= n; i++) { | ||
| map.put(br.readLine(), i); | ||
| } |
Collaborator
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아 해쉬맵으로 푸는 문제였어요? 짱짱bbb
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| Arrays.sort(ab); // 정렬 | ||
| Arrays.sort(cd); | ||
| long count = 0; | ||
| int start = 0; //ab의 시작지점 | ||
| int last = cd.length - 1; // cd의 시작지점 | ||
| while (start < n * n && last >= 0) { | ||
| int a = ab[start]; // 가장 작은 | ||
| int b = cd[last]; // 가장 큰 | ||
| if (a + b > 0) { // 합이 0보다 크면 | ||
| last--; // 큰 값 낮추기 | ||
| } else if (a + b < 0) { | ||
| start++; // 작은 값 낮추기 | ||
| } else { // 같으면 | ||
| int aCount = 0; | ||
| for (int i = start; i < ab.length; i++) { | ||
| if (a == ab[i]) { | ||
| aCount++; | ||
| } else { | ||
| break; | ||
| } | ||
| } | ||
| int bCount = 0; | ||
| for (int i = last; i >= 0; i--) { | ||
| if (b == cd[i]) { | ||
| bCount++; | ||
| } else { | ||
| break; | ||
| } | ||
| } |
Collaborator
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아... 이래서 정렬에 투포인터.. 배워갑니다 역시 갓상일,,
| System.out.println(sdp[idx]); | ||
| } | ||
| } | ||
| } |
eastsage
reviewed
Feb 28, 2023
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| if (tmp < dp[j]) { //만약 dp값이 더크면 | ||
| tmp = dp[j]; // tmp를 올려줌 | ||
| stmp = sdp[j]; // 문자열도 추가 | ||
| if (a[i].equals(b[j])){ // 만약 dp값을 설정해주었는데 같은 문자열이면 맥스값으로만 설정하고 넘어감 | ||
| continue; | ||
| } | ||
| } | ||
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| if (a[i].equals(b[j])) { // 만약 같은 문자열이면 | ||
| dp[j] = tmp + 1; // 현재 최고 값인 tmp를 현재 문자열 위치에 tmp+1 | ||
| sdp[j] = stmp + b[j]; //문자열도 현재 문자열 추가 | ||
| } |
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| System.out.println(count); // | ||
| } | ||
| } |
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주석달아놨으니 궁금한건 말로 설명하겠음...