kth-largest-element
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.
思路:
1.利用快速排序partition的思想
2.partition函数返回pivot的下标, 位于pivot左边的数都小于等于pivot,位于pivote右边的数都大于等于pivote.
3.当pivote + 1 == nums.length - k + 1 时,表明 nums[pivote] 为 从大到小的第K个的数
当 pivote + 1 < nums.length - k + 1 时, 表明 第k大的数在pivote 的右边
当 pivote + 1 > nums.length - k + 1 时,表明第K大的书在pivote的左边
class Solution { /* * @param k : description of k * @param nums : array of nums * @return: description of return */ public int kthLargestElement(int k, int[] nums) { // write your code here if (nums == null || nums.length == 0) { return 0; } if (k <= 0) { return 0; } return helper(nums, 0, nums.length - 1, nums.length - k + 1); } /* K 为小于等于第K大的数的个数 */ public int helper(int[] nums, int l, int r, int k) { if (l == r) { return nums[l]; } int position = partition(nums, l, r); if (position + 1 == k) { return nums[position]; } else if (position + 1 < k) { return helper(nums, position + 1, r, k); } else { return helper(nums, l, position - 1, k); } } public int partition(int[] nums, int l, int r) { if (l == r) { return l; } int left = l, right = r; int now = nums[left]; while (left < right) { while (left < right && nums[right] >= now) { right--; } nums[left] = nums[right]; while (left < right && nums[left] <= now) { left++; } nums[right] = nums[left]; } nums[left] = now; return left; } };