34. Search for a Range

最新推荐文章于 2022-01-21 16:09:17 发布

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最新推荐文章于 2022-01-21 16:09:17 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/qq_27655147/article/details/79389511

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本文介绍了一种算法，可以在已排序的整数数组中找到给定目标值的起始和结束位置，采用二分查找法实现O(log n)的时间复杂度。如果目标值不存在于数组中，则返回[-1, -1]。

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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].

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int min = 0, max = nums.size() - 1, mid = 0;
        vector <int>  res;
        int start = 0, end = 0;
        while(min <= max){
            mid = min + (max - min)/2;
            if(nums[mid] == target){
                start = mid;
                end = mid;
                while(start > 0 && nums[start] == nums[start - 1]){
                    --start;
                }
                while(end <= nums.size() - 1 && nums[end] == nums[end+1]){
                    ++end;
                }
                res.push_back(start);
                res.push_back(end);
                return res;
            }
            if(target < nums[mid]){
                max = mid - 1;
            }
            if(target > nums[mid]){
                min = mid + 1;
            }
        }
        res.push_back(-1);
        res.push_back(-1);
        return res;
    }
};