Description
Given an array of integers sorted in ascending order, 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].
my program
(1)时间复杂度为O（n）
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int i = 0, j = nums.size() -1;
vector<int> res;
while (i < nums.size())
{
if (nums[i] == target)
break;
i++;
}
while ( j>= 0)
{
if (nums[j] == target)
break;
j--;
}
if (i>=nums.size())
{
res.push_back(-1);
res.push_back(-1);
}else
{
res.push_back(i);
res.push_back(j);
}
return res;
}
};(2)时间复杂度为O（logn），利用二分查找
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int begin = 0, end = nums.size(), mid, left, right;
while (begin < end) {
mid = (begin + end) / 2;
if (nums[mid] >= target)
end = mid;
else
begin = mid + 1;
}
left = begin;
begin = 0, end = nums.size();
while (begin < end) {
mid = (begin + end) / 2;
if (nums[mid] > target)
end = mid;
else
begin = mid + 1;
}
right = begin;
return left == right ? vector<int> {-1,-1} : vector<int> {left,right-1};
}
};先找左边界。当mid >= target,将end移动到mid，否则（mid < target），begin = mid+1；
再找右边界。 当mid > target,将end移动到mid，否则（mid <= target），begin = mid+1；最后begin和end可能在target的左面一个。
如果没有数组中target的话，那么left和right会指到同一个值上，否则的话就意味着出现了target，此时left指向第一个出现的target，而right是指向最后一个target的下一个值，所以right需要减去1.