leetcode--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：给定一个排序数组，有重复元素。给定一个target，在数组中找到target的其实位置和结束位置

要求时间复杂度为O(logn)

如果target不在数组中，返回[-1,1];

分类：数组，二分法

解法1：二分查找找到target所在位置。然后从该位置开始，向两边查找边缘。

public class Solution {    public int[] searchRange(int[] nums, int target) {        int[] res = new int[2];    if(nums.length==0) return new int[]{-1,-1};    int low = 0;    int high = nums.length-1;    int mid = 0;        while(low<=high){    mid = (low+high)/2;    if(nums[mid]==target) break;    else if(nums[mid]>target) high = mid-1;    else low = mid+1;    }        if(nums[mid]==target){        int i = mid-1;        while(i>=0 && nums[i]==target){        i--;    }    if(nums[i+1]==target)    res[0] = i+1;    else    res[0] = mid;        i = mid+1;    while(i<nums.length&&nums[i]==target){        i++;    }    if(nums[i-1]==target)    res[1] = i-1;    else    res[1] = mid;           return res;    }else{    return new int[]{-1,-1};    }    }}