leetcode 二分查找 Search for a Range

Search for a Range

Total Accepted: 21480 Total
Submissions: 78454My Submissions

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].

题意：在一个排好序的数组，找给定值的起始下标和结束下标

思路：二分查找

先用lower_bound找到起始下标

再用upper_bound找到结束下标

lower_bound和upper_bound的实现参见《stl源码剖析》

复杂度：时间O(log n) 空间O(1)

vector<int> searchRange(int A[], int n, int target){
	int l = distance(A, lower_bound(A, A + n, target));
	int u = distance(A, upper_bound(A, A + n, target));
	vector<int> res;
	if(A[l] != target){
		res.push_back(-1);
		res.push_back(-1);
	}else{
		res.push_back(l);
		res.push_back(u - 1);
	}
	return res;
}
时间： 2024-08-06 03:14:04

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