[数据结构与算法]34（数组）. 在排序数组中查找元素的第一个和最后一个位置

Leetcode:34
题目描述：
给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

进阶：设计时间复杂度为O（log n）

拆分成左右边界
示例1：
右边界
left = 0,right = 5 ->mid = 0 + 2 = 2
left = 3,right = 5 ->mid = 3 + 1 = 4
left = 5(r_B),right = 5 ->mid = 5
left = 5(r_B),right = 4,跳出找到r_B

左边界
left = 0,right = 5 ->mid = 0 + 2 = 2
left = 3.right = 5 ->mid = 3+ 1 = 4
left = 3,right = 3(l_B)->mid = 3
left = 3,right = 2(L_B),跳出


即：L_B=2.r_B=5

class Solution {
    int[] searchRange(int[] nums, int target) {
        int leftBorder = getLeftBorder(nums, target);
        int rightBorder = getRightBorder(nums, target);
        // 情况一
        if (leftBorder == -2 || rightBorder == -2) return new int[]{-1, -1};
        // 情况三
        if (rightBorder - leftBorder > 1) return new int[]{leftBorder + 1, rightBorder - 1};
        // 情况二
        return new int[]{-1, -1};
    }

    int getRightBorder(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        int rightBorder = -2; // 记录一下rightBorder没有被赋值的情况
        while (left <= right) {
            int middle = left + ((right - left) / 2);
            if (nums[middle] > target) {
                right = middle - 1;
            } else { // 寻找右边界，nums[middle] == target的时候更新left
                left = middle + 1;
                rightBorder = left;
            }
        }
        return rightBorder;
    }

    int getLeftBorder(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        int leftBorder = -2; // 记录一下leftBorder没有被赋值的情况
        while (left <= right) {
            int middle = left + ((right - left) / 2);
            if (nums[middle] >= target) { // 寻找左边界，nums[middle] == target的时候更新right
                right = middle - 1;
                leftBorder = right;
            } else {
                left = middle + 1;
            }
        }
        return leftBorder;
    }
}