leetcode：Search for a Range 【Java】

一、问题描述

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].

二、问题分析

利用二分查找算法查找target下标，再结合计数器查找区间（注意：假设i > j，则i和j之间的元素个数count = i - j + 1）。

三、算法代码

public class Solution {    public int[] searchRange(int[] nums, int target) {        int [] result = new int[]{-1, -1};        int start = 0;        int end = nums.length - 1;        int middle = 0;        while(start <= end){        middle = (start + end)/2;        if(nums[middle] == target){        //找到target，开始查找区域        int count = 0;        for(int i = middle; i >= 0; i--){        if(nums[i] == target){        count++;        }else{        break;        }        }        result[0] = middle - count + 1;        count = 0;//重新初始化计数器        for(int j = middle; j <= nums.length - 1; j++){        if(nums[j] == target){        count++;        }else{        break;        }        }        result[1] = middle + count - 1;        return result;        }        if(nums[middle] > target){        end = middle - 1;        }else{        start = middle + 1;        }        }//end while                return result;    }}