Leetcode - Kth Largest Element in an Array

本文介绍两种有效的方法来找出未排序数组中的第K大元素。一种是通过排序数组直接获取,时间复杂度为O(nlogn);另一种是使用快速选择算法的递归版本,该方法更高效。同时提供了详细的算法实现代码。

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Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.

    // method 1: O(nlogn)
    public int findKthLargest1(int[] nums, int k) {
        Arrays.sort(nums);
        return nums[nums.length - k];
    }
    // method 2: quick select: standard recursive version
    public int findKthLargest2(int[] nums, int k) {
        quickSelect(nums, k, 0, nums.length - 1);
        return nums[nums.length - k];
    }
    private static final int CUTOFF = 10;
    public void quickSelect(int[] nums, int k, int left, int right) {
        if (left + CUTOFF <= right) {
            int pivot = median3(nums, left, right);
            int i = left, j = right - 1;
            while (true) {
                while(nums[++i] < pivot) {}
                while(nums[--j] > pivot) {}
                if (i < j) {
                    swap(nums, i, j);
                } else {
                    break;
                }
            }
            swap(nums, i, right - 1);
            if (nums.length - k < i)
                quickSelect(nums, left, i - 1, k);
            else if (nums.length - k > i)
                quickSelect(nums, i + 1, right, k);
        } else {
            insertionSort(nums, left, right);
        }
    }
    private int median3(int[] nums, int left, int right) {
        int mid = (left + right) / 2;
        if (nums[mid] < nums[left])
            swap(nums, left, mid);
        if (nums[mid] > nums[right])
            swap(nums, mid, right);
        if (nums[mid] < nums[left])
            swap(nums, left, mid);
        swap(nums, mid, right - 1);
        return nums[right - 1];
    }
    private void swap(int[] nums, int p1, int p2) {
        int tmp = nums[p1];
        nums[p1] = nums[p2];
        nums[p2] = tmp;
    }
    private void insertionSort(int[] nums, int left, int right) {
        for (int i = left + 1; i <= right; i++) {
            if (nums[i] < nums[i - 1]) {
                int tmp = nums[i];
                int j = i - 1;
                for (; j >= 0 && tmp < nums[j]; j--) {
                    nums[j + 1] = nums[j];
                }
                nums[j + 1] = tmp;
            }
        }
    }

 

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