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].
思路
二分查找的变形，不过可能存在数组元素重复的情况，所以在找到元素以后需要和已知的结果进行合并。如果之前没有结果，需要两边都进行查找；如果middle值大于之前的上界，只查找右边区域；如果middle值小于之前的下界，只查找左边区域；未找到的情况和一般的二分类似。代码如下。
代码
1 vector<int> result; 2 void initialize(){ 3 result.clear(); 4 result.push_back(-1); 5 result.push_back(-1); 6 } 7 void search(int A[], int left, int right, int target){ 8 if(left >= right) 9 return; 10 int middle = (left + right)/2; 11 if(A[middle] < target){ 12 search(A, middle+1, right, target); 13 } 14 else if(A[middle] > target){ 15 search(A, left, middle, target); 16 } 17 else{ 18 if(result[0] == -1){ 19 result[0] = middle; 20 result[1] = middle; 21 search(A, left, middle, target); 22 search(A, middle+1, right, target); 23 } 24 else{ 25 if(result[1] <= middle){ 26 result[1] = middle; 27 search(A, middle+1, right, target); 28 } 29 if(result[0] >= middle){ 30 result[0] = middle; 31 search(A, left, middle, target); 32 } 33 } 34 } 35 } 36 vector<int> searchRange(int A[], int n, int target) { 37 // Note: The Solution object is instantiated only once and is reused by each test case. 38 initialize(); 39 search(A, 0, n, target); 40 return result; 41 }