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]
给一串数字和一个目标数字,返回目标数字在这串数字中第一次出现的位置和最后出现的位置。注意这个目标数字是按升序排列的。
思路:这道题很简单,头尾两个指针分别查找就可以了。
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { vector<int> result={-1,-1}; int l=0, r=nums.size()-1; bool lFlag=false,rFlag=false; if(nums.empty()) return result; if(target<nums[0] || target>nums[nums.size()-1]) return result; while(l<=r){ if(nums[l]==target && !lFlag){ result[0]=l; lFlag=true; } if(nums[r]==target && !rFlag){ result[1]=r; rFlag=true; } if(lFlag&&rFlag) return result; if(target>nums[l]) l++; if(nums[r]>target) r--; } return result; } };