#21 22.07.24 > 정렬 > 두 배열의 원소 교체

beurmuz · Jul 24, 2022 · e40490d · e40490d
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diff --git a/src/cote/README.md b/src/cote/README.md
@@ -4,4 +4,4 @@
 |1|[그리디](./greedy/)|[거스름돈](./greedy/greedy_ex.js)<br>[큰 수의 법칙](./greedy/greedy_01.js)<br>[숫자 카드 게임](./greedy/greedy_02.js)<br>[1이 될 때까지](./greedy/greedy_03.js)<br>-<br>[모험가 길드](./greedy/test_01.js)<br>[곱하기 혹은 더하기](./greedy/test_02.js)<br>[문자열 뒤집기](./greedy/test_03.js)<br>|22.07.14<br><br><br><br>-<br>22.07.16<br><br><br>|O<br>X<br>O<br>O<br>-<br>O<br>O<br>O<br>|X<br>[22.07.16](./greedy/replay_01.js)<br>X<br>X<br>-<br>O<br>X<br>X<br>|
 |2|[구현](./implementation/)|[상하좌우](./implementation/imple_ex01.js)<br>[시각](./implementation/imple_ex02.js)<br>[왕실의 나이트](./implementation/imple_01.js)<br>[게임 개발](./implementation/imple_02.js)<br>-<br>[럭키 스트레이트](./implementation/test_01.js)<br>[문자열 재정렬](./implementation/test_02.js)<br>|22.07.15<br><br><br><br>-<br>22.07.20<br>22.07.23<br>|△<br>△<br>△<br>X<br>-<br>O<br>O|[22.07.17](./implementation/../replay_01.js)<br>[22.07.18](./implementation/replay_02.js)<br>[22.07.19](./implementation//replay_03.js)<br>[22.07.19](./implementation/replay_04.js)<br>-<br>X<br>X|
 |3|[DFS, BFS](./dfs%2Cbfs/)|[DFS 예제](./dfs%2Cbfs/dfs_ex.js)<br>[BFS 예제](./dfs%2Cbfs/dfs_ex.js)<br>[음료수 얼려 먹기](./dfs%2Cbfs/dfs%2Cbfs_01.js)<br>[미로 탈출](./dfs%2Cbfs/dfs%2Cbfs_02.js)<br>-<br>응|22.07.15<br><br>22.07.21<br>22.07.22<br>-<br>응|O<br>O<br>X<br>X<br>-<br>응|O<br>O<br>O<br>O<br>-<br>ㅇ|
-|4|[정렬](./sort/)|[위에서 아래로](./sort/sort_ex_01.js)<br>[성적이 낮은 순서로 학생 출력하기](./sort/sort_ex_02.js)<br>|22.07.22<br>22.07.24<br>|O<br>O<br>|X<br>X<br>|
+|4|[정렬](./sort/)|[위에서 아래로](./sort/sort_ex_01.js)<br>[성적이 낮은 순서로 학생 출력하기](./sort/sort_ex_02.js)<br>[두 배열의 원소 교체](./sort/sort_ex_03.js)|22.07.22<br>22.07.24<br><br>|O<br>O<br>O<br>|X<br>X<br>X<br>|
diff --git a/src/cote/sort/sort_ex_03.js b/src/cote/sort/sort_ex_03.js
@@ -0,0 +1,15 @@
+'use strict';
+
+function solution(N, K, s) {
+    const arr = s.split('\n').map((v) => v.split(' '));
+    const A = arr[0].map((v) => +v).sort((a,b) => a-b);
+    const B = arr[1].map((v) => +v).sort((a,b) => b-a);
+
+    console.log(A, B);
+    for(let i = 0; i < K; i++) {
+        [[A[i]], B[i]] = [[B[i]], A[i]];
+    }
+    return A.reduce((acc, value) => acc += value, 0);
+}
+
+console.log(solution(5, 3, '1 2 5 4 3\n5 5 6 6 5'));