LeetCode:Search for a Range

问题描述：

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：

1、利用二分查找法找到目标值；

2、从中间值向左找，找到与目标值相等的第一个值得下标，同理，找到右边最后一个相同值得下标；

例如： 3，4，5，6，6，6，6，6，6，7，8，9，左下标为3，右下标为8.

3、返回两个下标。

代码：

class Solution {
public:
int searchtarget(vector<int>& nums, int start,int end,int target)
{
int mid;
if(start > end) return -1;
else
{
mid = start + (end - start) / 2;
if(nums[mid] == target)
{
return mid;
}
else if(nums[mid] > target)
{
return searchtarget(nums,start,mid - 1,target);
}
else
{
return searchtarget(nums,mid + 1,end,target);
}
}
}
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> result;
int n = nums.size();
int index = searchtarget(nums,0,n-1,target);
if(index == -1)
{
result.push_back(-1);
result.push_back(-1);
return result;
}
else
{
int ls = index;
while(ls>0 && nums[index] == nums[ls-1])
<span style="white-space:pre">	</span>ls--;
int rs = index;
while(rs<n-1 && nums[index] == nums[rs+1])
<span style="white-space:pre">	</span>rs++;
result.clear();
result.push_back(ls);
result.push_back(rs);
}
return result;
}
};