Given a sorted array of integers, 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]
.
在有序数组中找出指定元素存在的区间。
两次二分,找到等于target和大于target的位置就行了
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { vector<int> result(2,-1); int l=lower_bound(nums.begin(),nums.end(),target)-nums.begin(); int r=upper_bound(nums.begin(),nums.end(),target)-nums.begin(); if(l!=r) { result[0]=l; result[1]=r-1; } return result; } };
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