Search for a Range

最新推荐文章于 2022-03-08 23:18:06 发布

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本文介绍如何在一个已排序的整数数组中找到给定目标值的起始和结束位置，确保算法的时间复杂度为O(log n)。如果目标值不存在于数组中，则返回[-1, -1]。

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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].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int min_index = -1;
        int max_index = -1;
        bool first_meet = true;
        for (int i = 0; i < n-1; ++i) {
            if (A[i] == target){
                if (first_meet) {
                   min_index = i;
                   first_meet = false;
                }
                if (A[i+1] > A[i]) {
                    max_index = i;
                }
            }
            
        }
        if (A[n-1] == target) {
            if (first_meet) {
               min_index = n - 1;
            }
            max_index = n - 1;
        }
        return  vector<int>({min_index, max_index});
        
    }
};