【LeetCode】34. 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]

.

由于O(logn)时间要求，显然用二分查找。

思路是先用二分查找找到其中一个target，找不到则返回默认的ret值[-1, -1]

找到之后从这个位置往两边递归进行二分查找进行范围的拓展。

具体来说，ret[0]不断向左扩展，ret[1]不断向右扩展。

class Solution {
public:
vector<int> searchRange(int A[], int n, int target) {
vector<int> ret(2, -1);
int left;
int right;
int low = 0;
int high = n-1;
while((left = binarySearch(A, low, high, target)) != -1)
{
ret[0] = left;
high = left-1;
}
low = 0;
high = n-1;
while((right = binarySearch(A, low, high, target)) != -1)
{
ret[1] = right;
low = right+1;
}
return ret;
}
int binarySearch(int A[], int left, int right, int target)
{
while(left <= right)
{
int mid = left + (right-left) / 2;
if(A[mid] == target)
return mid;
else if(A[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
return -1;
}
};