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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].
这题比较简单，就是边界的时候范围需要注意一下，我的方法先用二分法找到那个目标所在的位置，然后从那个位置增加和减少index直到不等于目标，于是就得到一个索引范围。代码如下：
1 class Solution { 2 public: 3 vector<int> searchRange(vector<int>& nums, int target) { 4 int len = nums.size(); 5 int start = 0; 6 int end = len; 7 vector<int> result; 8 findStartEnd(nums,start,end,target,result); 9 if(result.size()==0){ 10 result.push_back(-1); 11 result.push_back(-1); 12 } 13 return result; 14 } 15 16 void findStartEnd(vector<int>&nums,int start,int end,int target,vector<int>& result){ 17 int mid = (start+end)/2; 18 if(start<=mid&&end>=mid){ 19 if(nums[mid]>target){ 20 if(start == mid) mid = mid-2; 21 findStartEnd(nums,start,mid,target,result); 22 } 23 else if(nums[mid]<target){ 24 if(mid == end-1) mid = mid +2; 25 findStartEnd(nums,mid,end,target,result); 26 } 27 else{ 28 int front = mid,tail = mid; 29 for(int i = mid+1;i < end ;i++){ 30 if(nums[i]!=nums[mid]){ 31 tail = i-1; 32 break; 33 } 34 tail = i; 35 } 36 for(int j = mid-1 ;j>=start;j--){ 37 if(nums[j]!=nums[mid]){ 38 front = j+1; 39 break; 40 } 41 front = j; 42 } 43 result.push_back(front); 44 result.push_back(tail); 45 } 46 } 47 } 48 };
LeetCode 34 Search for a Range
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原文地址：http://www.cnblogs.com/yang-xiong/p/4970617.html
