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]
.
其实题意是想让自己写的二分,但是我想练习一下stl的二分的函数
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { vector<int> ans; if(lower_bound(nums.begin(),nums.end(),target) == nums.end()||*(lower_bound(nums.begin(),nums.end(),target)) != target) { ans.push_back(-1); ans.push_back(-1); return ans; } int k1 = lower_bound(nums.begin(),nums.end(),target) - nums.begin(); int k2 = upper_bound(nums.begin(),nums.end(),target) - nums.begin() -1; ans.push_back(k1); ans.push_back(k2); return ans; } };