算法分析与设计课程——LeetCode刷题之Search for a Range

题目：

Given an array of integers sorted in ascending order, 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]

.

答案：

vector<int> searchRange(vector<int>& nums, int target) {
int idx1 = lower_bound(nums, target);
int idx2 = lower_bound(nums, target+1)-1;
if (idx1 < nums.size() && nums[idx1] == target)
return {idx1, idx2};
else
return {-1, -1};
}

int lower_bound(vector<int>& nums, int target) {
int l = 0, r = nums.size()-1;
while (l <= r) {
int mid = (r-l)/2+l;
if (nums[mid] < target)
l = mid+1;
else
r = mid-1;
}
return l;
}