标签：排序 nbsp sed pos event complex not found 二分 tor
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]
找已排序数组中某个值的范围
1 class Solution { 2 public: 3 vector<int> searchRange(vector<int>& nums, int target) { 4 int size = nums.size(), s = 0, e = size - 1; 5 vector<int>ans; 6 if (size <= 0||target<nums[0]||target>nums[size-1]) { 7 ans.push_back(-1); 8 ans.push_back(-1); 9 return ans; 10 } 11 while (s < e) { 12 int mid = (s + e) / 2; 13 if (nums[mid] < target) 14 s = mid + 1; 15 else 16 e = mid; 17 } 18 if (nums[s] != target) { 19 ans.push_back(-1); 20 ans.push_back(-1); 21 return ans; 22 } 23 ans.push_back(s); 24 s = 0, e = size - 1; 25 while (s <= e) { 26 int mid = (s + e) / 2; 27 if (nums[mid] <= target) 28 s = mid + 1; 29 else 30 e = mid - 1; 31 } 32 ans.push_back(e); 33 return ans; 34 } 35 };
两次二分
19.2.2 [LeetCode 34] Find First and Last Position of Element in Sorted Array
标签：排序 nbsp sed pos event complex not found 二分 tor
原文地址：https://www.cnblogs.com/yalphait/p/10349393.html
