https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/
Given an array of integers nums 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].
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]
题解：二分法
代码：
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int idx = search(nums, 0, nums.size() - 1, target);
if (idx == -1) return {-1, -1};
int left = idx, right = idx;
while (left > 0 && nums[left - 1] == nums[idx]) left --;
while (right < nums.size() - 1 && nums[right + 1] == nums[idx]) right ++;
return {left, right};
}
int search(vector<int>& nums, int left, int right, int target) {
int mid;
while(left <= right) {
mid = (right - left) / 2 + left;
if(nums[mid] == target) return mid;
else if(nums[mid] < target) left = mid + 1;
else right = mid - 1;
}
return -1;
}
};