uploaded at 2023. 1. 5. 오후 7:01:57

programmers/가장 큰 수/solution.java

@@ -0,0 +1,60 @@
+// [문제 링크]: https://school.programmers.co.kr/learn/courses/30/lessons/42746#
+
+import java.util.*;
+import java.lang.StringBuilder;
+class Solution {
+    public String solution(int[] numbers) {
+        String answer = "";
+        Comp comp = new Comp();
+        StringBuilder sb = new StringBuilder();
+        String[] nb = new String[numbers.length];
+        for(int a = 0; a < numbers.length; a++){
+            nb[a] = String.valueOf(numbers[a]);
+        }
+        Arrays.sort(nb,comp);
+        for(String s : nb) sb.append(s);
+        return (sb.charAt(0)=='0')?"0":sb.toString();
+    }
+    public class Comp implements Comparator<String> {
+        public int compare(String a, String b) {
+            int temp = 0, end = Math.min(a.length(),b.length())-1;
+            int l = Integer.parseInt(a+b), r = Integer.parseInt(b+a);
+            while(true){
+                if(a.charAt(temp) < b.charAt(temp)) return 1;
+                else if(a.charAt(temp) > b.charAt(temp)) return -1;
+                if(temp < end) temp++;
+                else break;
+            } // 이거 통과헸다는 것은 동일 길이까지 전부 같다는 것.
+            if(a.length()==b.length()) return 0; // 이건 0 0 0 0 대비
+            if(a.length()>b.length()){
+                int tt = end+1, te = b.length(), mm = a.length();
+                // System.out.println(a+" "+b);
+                while(true){
+                    if(a.charAt(tt)>b.charAt(tt%te)) return -1;
+                    else if (a.charAt(tt)==b.charAt(tt%te)) {
+                        if(tt >= mm-1){
+                            if(l>r) return -1;
+                            else return 1;
+                        }
+                        else tt++;
+                    }
+                    else return 1; 
+                }
+                
+            } else {
+                int tt = end+1, te = a.length(), mm = b.length();
+                while(true){
+                    if(b.charAt(tt)>a.charAt(tt%te)) return 1;
+                    else if (b.charAt(tt)==a.charAt(tt%te)) {
+                        if(tt >= mm-1) {
+                            if(l<r) return 1;
+                            else return -1;
+                        }
+                        else tt++;
+                    }
+                    else return -1; 
+                }
+            }
+        }
+    }
+}