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;
}
};