LINQを使用した簡単な方法:
int[] top5 = array1.Concat(array2).Concat(array3).OrderByDescending(i => i).Take(5).ToArray();
最適な方法:
List<int> highests = new List<int>(); // Keep the current top 5 sorted
// Traverse each array. No need to put them together in an int[][]..it's just for simplicity
foreach (int[] array in new int[][] { array1, array2, array3 }) {
foreach (int i in array) {
int index = highests.BinarySearch(i); // where should i be?
if (highests.Count < 5) { // if not 5 yet, add anyway
if (index < 0) {
highests.Insert(~index, i);
} else { //add (duplicate)
highests.Insert(index, i);
}
}
else if (index < 0) { // not in top-5 yet, add
highests.Insert(~index, i);
highests.RemoveAt(0);
} else if (index > 0) { // already in top-5, add (duplicate)
highests.Insert(index, i);
highests.RemoveAt(0);
}
}
}
トップ5のソートされたリストを保持し、各配列を1回だけトラバースします。
BinarySearchを避けて、毎回トップ5の最低をチェックすることもできます。
List<int> highests = new List<int>();
foreach (int[] array in new int[][] { array1, array2, array3 }) {
foreach (int i in array) {
int index = highests.BinarySearch(i);
if (highests.Count < 5) { // if not 5 yet, add anyway
if (index < 0) {
highests.Insert(~index, i);
} else { //add (duplicate)
highests.Insert(index, i);
}
} else if (highests.First() < i) { // if larger than lowest top-5
if (index < 0) { // not in top-5 yet, add
highests.Insert(~index, i);
highests.RemoveAt(0);
} else { // already in top-5, add (duplicate)
highests.Insert(index, i);
highests.RemoveAt(0);
}
}
}
}