LeetCode题解： Search for a Range

Search for a Range

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].

思路：

二分搜索。可以参见 C++的 upper_bound和lower_bound的实现。

题解：

class Solution {
public:
    int lbound (int* A, int n, int target)
    {
        if (n == 0 || A == 0)
            return -1;
            
        int lower = 0;
        int upper = n - 1;
        
        while (lower < upper)
        {
            int mid = (lower + upper) / 2;
            if (A[mid] >= target)
                upper = mid;
            else if (A[mid] < target)
                lower = mid + 1;
        }
        
        if (A[lower] != target)
            return -1;
            
        return lower;
    }
    
    int ubound (int* A, int n, int target)
    {
        if (n == 0 || A == 0)
            return -1;
            
        int lower = 0;
        int upper = n - 1;
        
        while (lower < upper)
        {
            int mid = (lower + upper + 1) / 2;
            if (A[mid] > target)
                upper = mid - 1;
            else if (A[mid] <= target)
                lower = mid;
        }
        
        if (A[upper] != target)
            return -1;
        return upper;
    }

    vector<int> searchRange(int A[], int n, int target) {
        return vector<int>({{lbound(A, n, target), ubound(A, n, target)}});
    }
};