[LeetCode]Search for a Range

Search for a Range

My Submissions

Question

Total Accepted: 62397 Total Submissions: 225032 Difficulty: Medium

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].

Subscribe to see which companies asked this question

这个题目还是二分查找的变形，首先要找到一个mid使得nums[mid]==targer，从而左右再用二分查找开始寻找，

public class Solution {public int[] searchRange(int[] nums, int target) {        int [] result = new int[2];        result[0]=-1;result[1]=-1;        int head = 0;int tail = nums.length-1;        while(head<=tail){        int mid = head+(tail-head)/2;        if(nums[mid]==target){        int i=head;int j = mid;        while(i<j){        int mid1 = i+(j-i)/2;        if(nums[mid1]<target){        i=mid1+1;        }        else{        if(j==mid1)break;        j=mid1;        }        }        result[0]=j;        i=mid;j=tail;                while(i<j){        if(nums[j]==target){            i=j;            break;            }        int mid1 = i+(j-i)/2;        if(nums[mid1]>target){        j=mid1-1;        }        else{        if(i==mid1){        break;        }        i=mid1;        }        }        result[1]=i;        return result;        }        else{        if(nums[mid]>target){        tail=mid-1;        }        else{        head=mid+1;        }        }        }        return result;    }}