标签:
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]
.
1 class Solution { 2 public: 3 vector<int> searchRange(int A[], int n, int target) { 4 5 6 vector<int>res(2); 7 8 res[0]=bs(A,n,target-1)+1; 9 res[1]=bs(A,n,target); 10 11 if(res[1]==-1||A[res[1]]!=target) 12 { 13 res[0]=-1; 14 res[1]=-1; 15 } 16 return res; 17 } 18 19 //通过这个二分查找,如果有多个target的话可以找到最靠右边的元素 20 //同时也得注意,如果没有target则找到的是比target小的最大的最靠右元素 21 int bs(int A[],int n,int target) 22 { 23 int left=0; 24 int right=n-1; 25 int mid=(left+right)/2; 26 int ret=-1; 27 28 while(left<=right) 29 { 30 if(A[mid]>target) 31 { 32 right=mid-1; 33 } 34 else 35 { 36 //只要是当前元素小于等于target,left就会右移,因此找到最靠右的元素 37 ret=mid; 38 left=mid+1; 39 } 40 mid=(left+right)/2; 41 } 42 return ret; 43 } 44 45 };
标签:
原文地址:http://www.cnblogs.com/reachteam/p/4245958.html