最小的K个数

题目

牛客网

输入n个整数，找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字，则最小的4个数字是1,2,3,4,。

解题思路

Partition

该算法基于 Partition

  1. public ArrayList<Integer> GetLeastNumbers_Solution_Partition(int[] input, int k) {
  2.     ArrayList<Integer> res = new ArrayList<>();
  4.     if (k > input.length || k < 1) {
  5.         return res;
  6.     }
  8.     int start = 0, end = input.length - 1;
  9.     int index = partition(input, start, end);
  10.     while (index != k - 1) {
  11.         if (index > k - 1) {
  12.             end = index - 1;
  13.             index = partition(input, start, end);
  14.         } else {
  15.             start = index + 1;
  16.             index = partition(input, start, end);
  17.         }
  18.     }
  20.     for (int i = 0; i < input.length && i < k; i++) {
  21.         res.add(input[i]);
  22.     }
  23.     return res;
  24. }
  26. private int partition(int[] nums, int start, int end) {
  27.     int left = start, right = end;
  28.     int key = nums[left];
  30.     while (left < right) {
  31.         while (left < right && nums[right] > key) {
  32.             right--;
  33.         }
  34.         if (left < right) {
  35.             nums[left] = nums[right];
  36.             left++;
  37.         }
  39.         while (left < right && nums[left] <= key) {
  40.             left++;
  41.         }
  42.         if (left < right) {
  43.             nums[right] = nums[left];
  44.             right++;
  45.         }
  46.     }
  48.     nums[left] = key;
  50.     return left;
  51. }

小根堆算法

该算法基于小根堆，适合海量数据，时间复杂度为：n*logk

  1. public ArrayList<Integer> GetLeastNumbers_Solution(int[] input, int k) {
  2.     ArrayList<Integer> res = new ArrayList<>();
  3.     if (k > input.length||k==0) {
  4.         return res;
  5.     }
  7.     for (int i = input.length - 1; i >= 0; i--) {
  8.         minHeap(input, 0, i);
  10.         swap(input, 0, i);
  12.         res.add(input[i]);
  13.         if (res.size() == k) break;
  14.     }
  15.     return res;
  16. }
  18. private void minHeap(int[] heap, int start, int end) {
  19.     if (start == end) {
  20.         return;
  21.     }
  23.     int childLeft = start * 2 + 1;
  24.     int childRight = childLeft + 1;
  26.     if (childLeft <= end) {
  27.         minHeap(heap, childLeft, end);
  29.         if (heap[childLeft] < heap[start]) {
  30.             swap(heap, start, childLeft);
  31.         }
  32.     }
  34.     if (childRight <= end) {
  35.         minHeap(heap, childRight, end);
  37.         if (heap[childRight] < heap[start]) {
  38.             swap(heap, start, childRight);
  39.         }
  40.     }
  41. }
  43. private void swap(int[] nums, int a, int b) {
  44.     int t = nums[a];
  45.     nums[a] = nums[b];
  46.     nums[b] = t;
  47. }