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快速选择可以在nlogn的时间复杂度下得到一组无序数字的第k大,原理基于快速排序,每次以某个数字为中点分成的左右两段,左边都小于等于分割数,而右边都大于等于分割数,快速选择的求解速度要比全部排序后再取第k大快
/** * Created by tcgogogo on 16/8/18. */ class Solution { public void outPut(int[] nums) { for (int i = 0; i < nums.length; i++) { System.out.print(nums[i] + " "); } System.out.println(); } public void swap(int[] nums, int i, int j) { if (i == j) { return; } int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp; } public void quickSort(int[] nums, int left, int right) { if (left > right) { return; } int i = left, j = right; while (i < j) { while (i < j && nums[j] >= nums[left]) { j --; } while (i < j && nums[i] <= nums[left]) { i ++; } if(i < j) { swap(nums, i, j); } } swap(nums, left, i); quickSort(nums, left, i - 1); quickSort(nums, i + 1, right); } public int quickSelect(int[] nums, int left, int right, int k) { if (left > right) { return 0; } int i = left, j = right; while (i < j) { while (i < j && nums[j] >= nums[left]) { j --; } while (i < j && nums[i] <= nums[left]) { i ++; } swap(nums, i, j); } swap(nums, left, i); if(k == i - left + 1) { return nums[i]; } else if(k < i - left + 1) { return quickSelect(nums, left, i - 1, k); } else { return quickSelect(nums, i + 1, right, k - (i - left + 1)); } } }
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原文地址:http://blog.csdn.net/tc_to_top/article/details/52240331