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].
方法一：
参见STL lower_bound 和 upper_bound http://www.cnblogs.com/diegodu/p/3795077.html
，STL的equal_range就是基于lower_bound 和 upper_bound实现的，这里我们也可以利用这两个函数。。
注意没有找到目标时的判断，low == n表示所有元素都比targe大，A[low] != target且low！=n表示所有元素都比target小。。
1 int lower_bound(int* array, int low, int high, int key ) 2 { 3 int mid = 0; 4 while(low < high) 5 { 6 mid = (low + high)/2; 7 //cout << "low :" << low <<endl; 8 //cout << "high:" << high<<endl; 9 //cout << "mid:" << mid<<endl; 10 if(array[mid] >= key) 11 high = mid; 12 else 13 low = mid+1; 14 15 } 16 return high; 17 18 } 19 20 int upper_bound(int* array, int low, int high, int key ) 21 { 22 int mid = 0; 23 while(low < high) 24 { 25 mid = (low + high)/2; 26 if(array[mid] > key) 27 high = mid; 28 else 29 low = mid+1; 30 31 } 32 return high; 33 34 } 35 36 37 class Solution { 38 public: 39 vector<int> searchRange(int A[], int n, int target) { 40 41 int low = lower_bound(A, 0, n, target); 42 int high = upper_bound(A, 0, n, target); 43 vector<int> result; 44 45 if(low == n || A[low] != target) 46 { 47 result.push_back(-1); 48 result.push_back(-1); 49 } 50 else 51 { 52 result.push_back(low); 53 result.push_back(high-1); 54 } 55 56 return result; 57 } 58 };
方法二：
利用二分法：
1 class Solution { 2 public: 3 int m_low; 4 int m_high; 5 void search(int* array, int low, int high, int key ) 6 { 7 int mid = 0; 8 9 while(low <= high) 10 { 11 mid = (low + high)/2; 12 13 if(array[mid] == key) 14 { 15 if(mid < m_low) 16 m_low = mid; 17 if(mid > m_high) 18 m_high = mid; 19 search(array, low, mid-1, key); 20 search(array, mid+1, high, key); 21 return; 22 } 23 else if(array[mid] > key) 24 25 high = mid - 1; 26 else 27 low = mid+1; 28 29 } 30 31 } 32 vector<int> searchRange(int A[], int n, int target) { 33 34 vector<int> result; 35 36 m_low = INT_MAX; 37 m_high = INT_MIN; 38 39 search(A, 0, n-1, target); 40 41 if(m_low == INT_MAX) 42 { 43 result.push_back(-1); 44 result.push_back(-1); 45 } 46 else 47 { 48 result.push_back(m_low); 49 result.push_back(m_high); 50 } 51 52 return result; 53 } 54 };