美团2020，4.16笔试第三题

作者：左程云
链接：https://www.nowcoder.com/discuss/410116?type=0&order=0&pos=21&page=1
来源：牛客网

import java.util.Arrays;
import java.util.Comparator;

public class Code01_KthMinPair {

 public static class Pair {
  public int x;
  public int y;

  Pair(int x, int y) {
   this.x = x;
   this.y = y;
  }
 }

 public static class PairComparator implements Comparator {

  @Override
  public int compare(Pair arg0, Pair arg1) {
   return arg0.x != arg1.x ? arg0.x - arg1.x : arg0.y - arg1.y;
  }

 }

 // O(N^2)的复杂度，你肯定过不了
 public static int[] kthMinPair1(int[] arr, int k) {
  int N = arr.length;
  if (k > N * N) {
   return null;
  }
  Pair[] pairs = new Pair[N * N];
  int index = 0;
  for (int i = 0; i < N; i++) {
   for (int j = 0; j < N; j++) {
    pairs[index++] = new Pair(arr[i], arr[j]);
   }
  }
  Arrays.sort(pairs, new PairComparator());
  return new int[] { pairs[k - 1].x, pairs[k - 1].y };
 }



 作者：左程云
链接：https://www.nowcoder.com/discuss/410116?type=0&order=0&pos=21&page=1
来源：牛客网

 public static int[] kthMinPair2(int[] arr, int k) {
  int N = arr.length;
  if (k > N * N) {
   return null;
  }
  Arrays.sort(arr);
  int fristIndex = (k - 1) / N;
  int fristNum = arr[fristIndex];
  int fristNumLeftIndex = fristIndex;
  while (fristNumLeftIndex >= 0) {
   if (arr[fristNumLeftIndex] == fristNum) {
    fristNumLeftIndex--;
   } else {
    break;
   }
  }
  fristNumLeftIndex++;
  int fristNumRightIndex = fristIndex;
  while (fristNumRightIndex < N) {
   if (arr[fristNumRightIndex] == fristNum) {
    fristNumRightIndex++;
   } else {
    break;
   }
  }
  fristNumRightIndex--;
  int restTh = k - (fristNumLeftIndex * N);
  return new int[] { fristNum, arr[(restTh - 1) / (fristNumRightIndex - fristNumLeftIndex + 1)] };
 }