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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]
1 class Solution { 2 public: 3 vector<int> searchRange(vector<int>& nums, int target) { 4 int len=nums.size(); 5 vector<int> res; 6 int flag=0; 7 for(int i=0;i<len;i++) 8 { 9 if(target==nums[i]) 10 { 11 res.push_back(i); 12 flag=1; 13 break; 14 } 15 } 16 17 if(flag==0) 18 { 19 res.push_back(-1); 20 } 21 flag=0; 22 for(int i=len-1;i>=0;i--) 23 { 24 if(target==nums[i]) 25 { 26 res.push_back(i); 27 flag=1; 28 break; 29 } 30 } 31 if(flag==0) 32 { 33 res.push_back(-1); 34 } 35 36 37 return res; 38 } 39 };
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原文地址：http://www.cnblogs.com/jawiezhu/p/4510162.html
