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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].
思路
通过二分查找，如果没找到返回[-1,-1]如果找到middle，找出start和ending即可
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 boolean found = false; 4 int low = 0; 5 int high = A.length - 1; 6 int middle = 0; 7 int result[] = new int[2]; 8 9 while(low <= high){10 middle = (low + high) / 2;11 if(target > A[middle]){12 low = middle + 1;13 }14 else if(target < A[middle]){15 high = middle - 1;16 }17 else18 {19 found = true;20 break;21 }22 }//while23 if(!found) //没有查找到24 { 25 result[0] = -1;26 result[1] = -1;27 }//if28 else{29 while(middle >= 0 && A[middle] == target){30 middle--;31 }32 if(middle < 0)33 result[0] = 0;34 else35 result[0] = middle + 1;36 middle = result[0];37 while(middle < A.length && A[middle] == target){38 middle++;39 }40 if(middle >= A.length)41 result[1] = A.length - 1;42 else43 result[1] = middle - 1;44 }//else45 return result;46 }47 }
Search for a Range
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