Your algorithm's runtime complexity must be in the order of O(log n).
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].
#include<iostream>
#include<vector>
using namespace std;
vector<int> searchRange(int A[], int n, int target) {
int first = 0;
int last = n - 1;
vector<int>result(2, -1);
while (first<=last)
{
int mid = (first + last) / 2;
if (A[mid] == target)
{
result[0] = mid;
result[1] = mid;
while (result[0]-1 >= first&&A[result[0]-1] == target)//当一位是反复位时才对范围跟新
--result[0];
while (result[1]+1 <= last&&A[result[1]+1] == target)
++result[1];
return result;
}
else if (A[mid] < target)
first = mid + 1;
else
last = mid - 1;
}
return result;
}
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