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Search for a Range
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].
Analyse: binary search.
1 class Solution { 2 public: 3 vector<int> searchRange(vector<int>& nums, int target) { 4 vector<int> range = {-1, -1}; 5 int left = 0, right = nums.size() - 1; 6 while (left <= right) { 7 int mid = left + (right - left) / 2; 8 if (nums[mid] > target) right = mid - 1; 9 else if (nums[mid] < target) left = mid + 1;10 else {11 int start = mid, end = mid;12 while (start >= 0 && nums[start] == nums[mid]) start--;13 while (end < nums.size() && nums[end] == nums[mid]) end++;14 range[0] = start + 1;15 range[1] = end - 1;16 break;17 }18 }19 return range;20 }21 };
Search for a Range
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