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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].
Solution:
本题时间复杂度是O(log n)，暗示我们应该采用binary search的方法来做此题。
关键点在于有重复的元素，所以需要利用findMostLeft和findMostRight来找到起始和结束的位置。
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int pos=binarySearch(A,target); 4 5 if(pos==-1){ 6 return new int[] {-1,-1}; 7 }else{ 8 int left=-1; 9 int right=-1; 10 if(pos-1>=0&&A[pos]==A[pos-1]){ 11 left=findMostLeft(A,target,0,pos-1); 12 } 13 else{ 14 left=pos; 15 } 16 if(pos+1<=A.length-1&&A[pos]==A[pos+1]){ 17 right=findMostRight(A,target,pos+1,A.length-1); 18 19 }else{ 20 right=pos; 21 } 22 return new int[] {left,right}; 23 } 24 } 25 public int findMostRight(int[] A,int target,int left,int right){ 26 27 while(left<=right){ 28 int mid=left+(right-left)/2; 29 if(A[mid]==target){ 30 if(mid+1<=right&&A[mid]==A[mid+1]) 31 left=mid+1; 32 else 33 return mid; 34 }else{ 35 right=mid-1; 36 } 37 } 38 return -1; 39 } 40 public int findMostLeft(int[] A,int target,int left,int right){ 41 42 43 while(left<=right){ 44 int mid=left+(right-left)/2; 45 if(A[mid]==target){ 46 if(mid-1>=left&&A[mid]==A[mid-1]) 47 right=mid-1; 48 else 49 return mid; 50 }else{ 51 left=mid+1; 52 } 53 mid=left+(right-left)/2; 54 } 55 return -1; 56 } 57 public int binarySearch(int[] A, int target){ 58 int left=0; 59 int right=A.length-1; 60 int mid=left+(right-left)/2; 61 while(left<=right){ 62 63 64 if(A[mid]==target){ 65 66 return mid; 67 }else if(A[mid]<target){ 68 left=mid+1; 69 }else{ 70 right=mid-1; 71 } 72 mid=left+(right-left)/2; 73 } 74 return -1; 75 } 76 }
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原文地址：http://www.cnblogs.com/Phoebe815/p/4261430.html
