Leetcode - Kth Largest Element in an Array

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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转载自likesky3.iteye.com/blog/2213761