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].
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Analysis:
We need to do two binary search. The first finds out the left boundary and the second finds out the right boundary.
if mid==target, we then check whether mid-1==target, if yes, we then continue to search [start,mid-1] until we find the left boundary.
Solution:
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int[] res = new int[]{-1,-1}; 4 if (A.length==0) return res; 5 6 int start = 0, end = A.length-1; 7 8 //Find left range. 9 10 while (start<=end){ 11 int mid = (start+end)/2; 12 if (A[mid]==target){ 13 //find right range also. 14 if (mid+1==A.length || A[mid+1]!=target) res[1] = mid; 15 16 //Check left 17 if (mid-1==-1 || A[mid-1]!=target){ 18 res[0]=mid; 19 break; 20 } else { 21 end = mid-1; 22 continue; 23 } 24 } else if (A[mid]>target) end = mid-1; 25 else start = mid+1; 26 } 27 28 if (start>end) return res; 29 if (res[0]!=-1 && res[1]!=-1) return res; 30 31 //Find right 32 start = 0; 33 end = A.length-1; 34 while (start<=end){ 35 int mid = (start+end)/2; 36 if (A[mid]==target){ 37 //Check right 38 if (mid+1==A.length || A[mid+1]!=target){ 39 res[1]=mid; 40 break; 41 } else { 42 start = mid+1; 43 continue; 44 } 45 } else if (A[mid]>target) end = mid-1; 46 else start = mid+1; 47 } 48 49 return res; 50 } 51 }