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].

 

A solution of o(logn). Do two runs of binary search for the index of largest element strictly less than target, and the index of largest element strictly less than target+1. The range is between this two indices. 

 1 public class Solution {

 2     public int[] searchRange(int[] A, int target) {

 3         // Start typing your Java solution below

 4         // DO NOT write main() function

 5         if(A==null || A.length==0)return null;

 6         int[] res = new int[2];

 7         res[0]=searchMaxLessThan(A,target,0,A.length-1);

 8         res[1]=searchMaxLessThan(A,target+1,0,A.length-1);

 9         if(res[0]==res[1]){

10             res[0]=-1;

11             res[1]=-1;

12         }else{

13             res[0]++;

14         }

15         return res;

16     }

17     

18     public int searchMaxLessThan(int[] A, int target, int start, int end){

19         

20         if(start==end) return A[start]<target?start:start-1;

21         if(start==end-1) return A[end]<target?end:(A[start]<target?start:start-1);

22         int mid = (start+end)/2;

23         if(A[mid]>=target){

24             end=mid-1;

25         }else{

26             start=mid;

27         }

28         return searchMaxLessThan(A,target,start,end);

29     }

30 }

 第二遍：

很巧妙的采用两次循环来做。

 1 public class Solution {

 2     public int[] searchRange(int[] A, int target) {

 3         // Start typing your Java solution below

 4         // DO NOT write main() function

 5         int start = 0;

 6         int end = A.length - 1;

 7         int l = -1;

 8         int r = -1;

 9         while(start <= end){

10             int mid = (start + end) / 2;

11             if(A[mid] == target) l = mid;

12             if(target <= A[mid]) end = mid - 1;

13             else start = mid + 1;

14         }

15         start = 0;

16         end = A.length - 1;

17         while(start <= end){

18             int mid = (start + end) / 2;

19             if(A[mid] == target) r = mid;

20             if(target >= A[mid]) start = mid + 1;

21             else end = mid - 1;

22         }

23         return new int[]{l, r};

24     }

25 }

 第三遍：

 1 public class Solution {

 2     public int[] searchRange(int[] A, int target) {

 3         int len = A.length;

 4         int srt = 0, end = len - 1;

 5         int left = -1, right = -1;

 6         while(srt <= end){

 7             int mid = (srt + end) / 2;

 8             if(A[mid] < target) srt = mid + 1;

 9             else if(A[mid] > target) end = mid - 1;

10             else{

11                 left = mid;

12                 end = mid - 1;

13             }

14         }

15         srt = 0; end = len - 1;

16         while(srt <= end){

17             int mid = (srt + end) / 2;

18             if(A[mid] < target) srt = mid + 1;

19             else if(A[mid] > target) end = mid - 1;

20             else{

21                 right = mid;

22                 srt = mid + 1;

23             }

24         }

25         return new int[]{left, right};

26     }

27 }