https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.
If target is not found in the array, return [-1, -1].
You must write an algorithm with O(log n) runtime complexity.
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
Example 3:
Input: nums = [], target = 0
Output: [-1,-1]
Constraints:
0 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
nums is a non-decreasing array.
-10^9 <= target <= 10^9
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
if (nums.size()==0) return {-1,-1};
int left=0, right=nums.size()-1;
int middle=(right-left)/2+left;
while (left<=right){
middle=(right-left)/2+left;
if (nums[middle]==target) break;
if (nums[middle]<target) left=middle+1;
else right=middle-1;
}
if (nums[middle]!=target) return {-1,-1};
left=middle;right=middle;
while (left>=0 && nums[left]==target) left--;
while (right<nums.size() && nums[right]==target) right++;
return {left+1,right-1};
}
};