LeetCode 0034 - Search for a Range

Reck Zhang

发布于 2021-08-11 10:23:23

1800

发布于 2021-08-11 10:23:23

文章被收录于专栏：Reck Zhang

Search for a Range

Desicription

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].

Solution

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res;
        auto tmp = lower_bound(nums.begin(), nums.end(), target);
        if(tmp == nums.end() || *tmp != target){
            res.push_back(-1);
            res.push_back(-1);
            return res;
        }
        res.push_back(tmp-nums.begin());
        tmp = upper_bound(nums.begin(), nums.end(), target);
        res.push_back(tmp-nums.begin()-1);
        return res;
    }
};

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原始发表：2017-11-11，如有侵权请联系 cloudcommunity@tencent.com 删除

algorithm