Leetcode34. 在排序数组中查找元素的第一个和最后一个位置

Leetcode34. 在排序数组中查找元素的第一个和最后一个位置

题目：
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。
你的算法时间复杂度必须是 O(log n) 级别。
如果数组中不存在目标值，返回 [-1, -1]。
示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]

示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]

题解：
二分法，查找左边界和右边界

java代码：

 public static int[] searchRange(int[] nums, int target) {
        int[] res = new int[2];
        int left = 0;
        int right = nums.length - 1;

        //找到右边界
        while (left <= right) {
            int mid = (right - left) / 2 + left;
            if (nums[mid] > target) {
                right = mid - 1;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                if (mid == nums.length - 1 || nums[mid + 1] > target) {
                    res[1] = mid;
                    break;
                }
                left = mid + 1;
            }
        }
        left = 0;
        right = nums.length - 1;

        //找到左边界
        while (left <= right) {
            int mid = (right - left) / 2 + left;
            if (nums[mid] > target) {
                right = mid - 1;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                if (mid == 0 || nums[mid - 1] < target) {
                    res[0] = mid;
                    break;
                }
                right = mid - 1;
            }
        }

        return res;
    }

scala代码：

/**
    *
    * @param nums
    * @param target
    * @return
    */
  def searchRange(nums: Array[Int], target: Int): Array[Int] = {

    val arr = new Array[Int](2)
    var left = 0
    var right = nums.length - 1
    //找到右边界
    breakable {
      while (left <= right) {
        val mid = left + (right - left) / 2
        if (nums(mid) > target) {
          right = mid - 1
        } else if (nums(mid) < target) {
          left = mid + 1
        } else {
          if (mid == nums.length - 1 || nums(mid + 1) > target) {
            arr(1) = mid
            break()
          } else {
            left = mid + 1
          }
        }
      }
    }

    left =0
    right =nums.length-1
    //找到左边界
    breakable {
      while (left <= right) {
        val mid = left + (right - left) / 2
        if (nums(mid) > target) {
          right = mid - 1
        } else if (nums(mid) < target) {
          left = mid + 1
        } else {
          if (mid == 0 || nums(mid - 1) < target) {
            arr(0) = mid
            break()
          } else {
            right = mid - 1
          }
        }
      }
    }
    arr
  }