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].
给的那个一个有序数组和一个目标target，找到这个目标在有序数组里的开始和结束的下标。如果不存在就返回 -1 -1。注，如果数组中目标值只有一个，那就返回两个相同的下标。
用binary search找到target 然后再在左边binary search找到左边（包括mid）第一个目标，记录下标。再在右边binary 找到最后一个目标。记录下标。返回。
代码如下：
class Solution {
public:
vector<int> searchRange(int A[], int n, int target)
{
vector<int> wr, ans;
wr.push_back(-1); // 要学会可以用 vector<int> wr(2,-1);直接将其初始化为想要的
wr.push_back(-1);
if(n == 0)
return wr;
int l = 0, r = n -1, mid = 0;
while(l <= r)
{
mid = (l+r)/2;
if (A[mid] < target) // 如果mid小于target那么可以把左边一块去掉
{
l = mid + 1;
continue;
}
if (A[mid] > target) // 如果mid大于target那么可以把右边一块去掉
{
r = mid -1;
continue;
}
if (A[mid] == target)// mid 找到了刚好，那就肯定在这里面要有return，如果一直没有找到，跳出while后就返回wr
{
// find start from l to mid，binary search
int start = l, l_r = mid; // l_r指mid左边的最右
while(l <= l_r)
{
int tmp_mid = (l + l_r)/2;
if(A[tmp_mid] != target)
{l = tmp_mid + 1;}
if(A[tmp_mid] == target)
{l_r = tmp_mid - 1; start = tmp_mid;}
}
ans.push_back(start);
// find end from mid to r，binary search
int r_l = mid, End = mid; // r_l指mid右边的最左
while(r_l <= r)
{
int tmp_mid = (r_l + r)/2;
if(A[tmp_mid] != target)
{r = tmp_mid - 1;}
if(A[tmp_mid] == target)
{r_l = tmp_mid + 1; End = tmp_mid;}
}
ans.push_back(End);
return ans;
}
}
return wr;
}
};