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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].
这道题类似上一道题，但是这道题多出了一个可能存在重复的条件，
对于这道题的解题思路为：首先用二分查找找出要找的元素所在的位置，然后向前或向后查找所有的要找的元素
1 package Search.fora.Range; 2 3 public class SearchforaRange { 4 5 /** 6 * @param args 7 */ 8 public static void main(String[] args) { 9 // TODO Auto-generated method stub 10 // int[] A={5, 7, 7, 8, 8, 10}; 11 int[] A={1,2,3}; 12 SearchforaRange service=new SearchforaRange(); 13 //int [] result=service.searchRange(A, 8); 14 int [] result=service.searchRange(A, 1); 15 System.out.println(result[0]); 16 System.out.println(result[1]); 17 } 18 public int[] searchRange(int[] A, int target) { 19 int [] result=new int[2]; 20 int len=A.length; 21 int low=0; 22 int hi=len-1; 23 while(low<=hi){ 24 int mid=(low+hi)/2; 25 if(A[mid]==target){ 26 //往后找点 27 int j=mid+1; 28 int i=mid-1; 29 while(j<len&&A[j]==A[mid]){ 30 j++; 31 } 32 j=j-1; 33 while(i>=0&&A[i]==A[mid]){ 34 i--; 35 } 36 i++; 37 result[0]=i; 38 result[1]=j; 39 return result; 40 } 41 if(A[mid]<A[low]){ 42 if(target>A[mid]&&target<=A[hi]){ 43 low=mid+1; 44 }else{ 45 hi=mid-1; 46 } 47 }else{ 48 if(A[low]<=target&&target<A[mid]){ 49 hi=mid-1; 50 }else{ 51 low=mid+1; 52 } 53 } 54 } 55 result[0]=-1; 56 result[1]=-1; 57 return result; 58 } 59 }
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原文地址：http://www.cnblogs.com/criseRabbit/p/4268841.html
