【LeetCode】91. Search for a Range

题目描述(Easy)

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 ofO(log n).

If the target is not found in the array, return[-1, -1].

题目链接

【牛客网】search-for-a-range 

Note: leetcode原题地址找不到了，如果有找到请私信我。

Example 1:

Given[5, 7, 7, 8, 8, 10]and target value 8,
return[3, 4].

算法分析

方法一：二分法+夹逼查找，最坏情况时间复杂度为；

方法二：二分法，先找到起始位，再找到末位。

提交代码：

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int lower = lower_bound(A, 0, n, target);
        int upper = upper_bound(A, lower, n, target);
        
        if (lower == n || A[lower] != target)
            return vector<int>{-1, -1};
        else
            return vector<int>{lower, upper - 1};
    }
    
private:    
    int lower_bound(int A[], int first, int last, int target) {
        while (first < last)
        {
            int mid = (first + last) / 2;
            
            if (target > A[mid]) first = mid + 1;
            else last = mid;
        }
        
        return first;
    }
    
    int upper_bound(int A[], int first, int last, int target) {
        while (first < last)
        {
            int mid = (first + last) / 2;
            
            if (target >= A[mid]) first = mid + 1;
            else last = mid;
        }
        
        return first;
    }
};