簡單說明：給定一個陣列(array)，該array裡面皆為數字組成，每個數字的位數間做總和，相同的成一組。每組至少兩個數字或以上，如果沒有，則不算成組，若全部都沒有成組，則回傳一個-1；若有成組，則將該組最大與次大的兩個數字做加總，比較其他組看誰的加總最大，便回傳該值。 example A=[51,71,17,42], the function should return 93. There are two pairs of numbers whose digits add up to an equal sum: (51, 42) and (17,71). The first pair sums up to 93. B=[42,33,60], the function should return 102. The digits of all numbers in A add up to the same sum, and choosing to add 42 and 60 gives the result 102. C=[51,32,43], the function should return -1, since all numbers in A have digits that add up to different, unique sums.
public static int solution(int[] A) { long startTime = System.nanoTime(); // 測試時間 HashMap<Integer, ArrayList<Integer>> entity = new HashMap<>(); ArrayList<Integer> list; for(int i=0 ; i<A.length ; i++) { int total = addAllDigit(A[i]); // 每一位數字相加 // 先找尋entity的key有沒有相同的 if(entity.containsKey(total)) { // 有相同 // 拿出原本map裡的list ArrayList<Integer> internal = entity.get(total); internal.add(A[i]); // 覆蓋存回該map裡面 entity.put(total, internal); } else { // 沒有相同 list = new ArrayList<>(); list.add(A[i]); entity.put(total, list); } } int biggest = -1; for(Integer value : entity.keySet()) { Integer array[] = (entity.get(value)).toArray(new Integer[0]); int total = addTwoNumberBiggest(array); if(total != -1 && total>biggest) { biggest = total; } } long endTime = System.nanoTime(); // 測試時間 System.out.println("spend Time :" + (endTime-startTime)/1000000); return biggest; }
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// 每一位數字相加 public static int addAllDigit(int number) { int flag=1; int result=0; result += number%10; number = number/10; while(flag != -1) { result += number%10; if((number/10) != 0) { number = number/10; } else { flag = -1; } } return result; }
