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;
    }
}