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) {
        int leftBorder = getLeftRange(weights, target);
        int rightBorder = getRightRange(weights, target);
        int[] result = new int[2];
        if (leftBorder == -2 || rightBorder == -2) {
            result[0] = -1;
            result[1] = -1;
            return result;
        }
        if (rightBorder - leftBorder > 1) {
            result[0] = leftBorder + 1;
            result[1] = rightBorder - 1;
            return result;
        }
        result[0] = -1;
        result[1] = -1;
        return result;
    }
    public int getRightRange(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int rightBorder = -2;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] >= target) {
                left = mid + 1;
                rightBorder = left;
            } else {
                right = mid - 1;
            }
        }
        return rightBorder;
    }

    public int getLeftRange(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int leftBorder = -2;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] <= target) {
                right = mid - 1;
                leftBorder = right;
            } else {
                left = mid + 1;
            }
        }
        return leftBorder;
    }
}

本题知识点分析：

1.二分查找

2.双指针

3.数组遍历和赋值

本题解题思路分析：

1.创建两个单独的功能函数用于寻找最左边界和最右边界

2.如果无法找到目标元素,返回-1,-1

3.如果rightBorder-leftBorder>1说明有连续区间，将left+1,right-1,然后进行返回

4.两个功能函数就是二分查找，无非是等号和什么时候取+1的问题，只要思考好自己找的好最左边还是最右边，是等于还是大于还是小于，这些问题就很容易做出

本题使用编程语言: Java