题解 | #找出特定体重的牛群#

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 target int整型
     * @return int整型一维数组
     */
    public int[] searchRange (int[] weights, int target) {
        // write code here
        int leftIndex = findLeftIndex(weights, target);
        if (leftIndex == -1) {
            return new int[] {-1, -1};
        }

        int rightIndex = findRightIndex(weights, target);
        return new int[] {leftIndex, rightIndex};
    }

    private int findLeftIndex(int[] weights, int target) {
        int left = 0, right = weights.length - 1;
        int index = -1;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (weights[mid] == target) {
                index = mid;
                right = mid - 1;
            } else if (weights[mid] < target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return index;
    }

    private int findRightIndex(int[] weights, int target) {
        int left = 0, right = weights.length - 1;
        int index = -1;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (weights[mid] == target) {
                index = mid;
                left = mid + 1;
            } else if (weights[mid] < target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return index;
    }
}

考察的知识点：

1. 二分查找：利用二分查找的思想在有序数组中寻找目标元素的起始和结束位置。
2. 数组操作：根据比较结果更新查找范围。

解题思路：

这个问题可以使用二分查找来实现，以达到时间复杂度为 O(log n)。首先，我们可以使用二分查找找到体重等于目标体重的任意一个牛的索引，然后在该索引的基础上向左和向右分别找到起始位置和结束位置。