LeetCode题解：Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

题意：给定有序数组，找到目标元素出现的起始位置和终止位置，若没有出现则返回-1，-1

解决思路：二分查找找到目标元素出现位置，然后找目标元素+1的数出现的元素的位置，减去1就是最后的结果了

代码：

public class Solution {    public int[] searchRange(int[] nums, int target) {        int left = binarySearch(nums, target);        if(left == nums.length || nums[left] != target){            return new int[]{-1, -1};        }        return new int[]{left, binarySearch(nums, target + 1) - 1};    }    private int binarySearch(int[] nums, int target){        int count = nums.length;        int step = 0;        int left = 0;        int mid = 0;        while(count > 0){            step = count >> 1;            mid = left + step;            if(nums[mid] < target){                left = mid + 1;                count -= (step + 1);            }else{                count = step;            }        }        return left;    }}