lintcode-easy-Median

最新推荐文章于 2020-12-27 04:00:25 发布

转载最新推荐文章于 2020-12-27 04:00:25 发布 · 64 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/goblinengineer/p/5234545.html

本文介绍了一种高效的方法来寻找未排序整数数组的中位数，通过使用快速选择算法直接定位到第 N/2 个元素。

Given a unsorted array with integers, find the median of it.

A median is the middle number of the array after it is sorted.

If there are even numbers in the array, return the N/2-th number after sorted.

Given [4, 5, 1, 2, 3], return 3.

Given [7, 9, 4, 5], return 5.

用quick select的方法直接找到第N/2个元素

public class Solution {
    /**
     * @param nums: A list of integers.
     * @return: An integer denotes the middle number of the array.
     */
    public int median(int[] nums) {
        // write your code here
        if(nums == null || nums.length == 0)
            return 0;
        
        int length = nums.length;
        
        return quickSelect(nums, 0, length - 1, (length - 1) / 2);
    }
    
    public int quickSelect(int[] nums, int start, int end, int k){
        int left = start;
        int right = end;
        int pivot = end;
        
        while(true){
            while(left < right && nums[left] < nums[pivot])
                left++;
            while(left < right && nums[right] >= nums[pivot])
                right--;
            
            if(left == right)
                break;
            else
                swap(nums, left, right);
        }
        
        swap(nums, left, pivot);
        
        if(left == k)
            return nums[left];
        else if(left < k)
            return quickSelect(nums, left + 1, end, k);
        else
            return quickSelect(nums, start, left - 1, k);
    }
    
    public void swap(int[]nums, int i, int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
        
        return;
    }
    
}

转载于:https://www.cnblogs.com/goblinengineer/p/5234545.html