题解 | #第k轻的牛牛#

知识点：快速排序

对于时间复杂度为 O(n) 的方法，我们需要使用快速排序的思想。

随机选择一个枢轴index，利用快速排序的方法，将大于该weights[index]的值移动到index右边，小于weights[index]的值移动到index左边，这样就做到了index位置的值一定是第index-1小的值。若index大于k-1，则说明第k小的值在左边，只需要递归左侧子数组即可，同理，若index小于k-1，说明第k小的值在右边，则递归右侧子数组。这样只需要遍历一半的空间，提高了时间效率。

Java题解如下：

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param weights int整型一维数组 
     * @param k int整型 
     * @return int整型
     */
    public int findKthSmallest (int[] weights, int k) {
        // write code here
        return quickSelect(weights, 0, weights.length - 1, k); 
    }

    private int quickSelect(int[] weights, int left, int right, int k) {
        Random random = new Random();
        int index = left + random.nextInt(right - left + 1);
        int target = weights[index];
        int i = left, j = right;
        weights[index] = weights[j];
        while(i < j) {
            while(i < j && weights[i] <= target) {
                i++;
            }
            weights[j] = weights[i];
            while(i < j && weights[j] >= target) {
                j--;
            }
            weights[i] = weights[j];
        }
        weights[j] = target;
        if(i == k - 1) {
            return weights[i];
        }else if(i > k - 1) {
            return quickSelect(weights, left, i - 1, k);
        }else {
            return quickSelect(weights, i + 1, right, k);
        }
    }
}