leetcode——34——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]

.

自己写的代码，没有真正的实现二分查找，运行速度相对较慢

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int i = 0;
int j = nums.size() - 1;
int mid;
vector<int> res;
while (i <= j)
{
mid = i + (j - i) / 2;
if (nums[mid] > target)
{
j = mid - 1;
}
else if (nums[mid] < target)
{
i = mid + 1;
}
else
{
int m = mid, n = mid;
while (nums[m] == target&&nums[m+1] ==target&&m<nums.size()-1)
{
m++;
}
while (nums
== target&&nums[n-1] == target&&n>0)
{
n--;
}
res.push_back(n);
res.push_back(m);
return res;
}
}
res.push_back(-1);
res.push_back(-1);
return res;
}
};

完全实现二分查找的代码

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res(2, -1);
if(nums.empty()) return res;
int high = binarySearchUp(nums, target, 0, nums.size() -1);
int low = binarySearchLow(nums, target, 0, nums.size() - 1);
if(high >= low)
{
res[0] = low;
res[1] = high;
return res;
}
return res;

}
private:
int binarySearchLow(vector<int>& nums, int target, int begin, int end)
{
if(begin > end) return begin;
int mid = begin + (end - begin) / 2;
if(nums[mid] < target) return binarySearchLow(nums, target, mid + 1, end);
else return binarySearchLow(nums, target, begin, mid - 1);
}
int binarySearchUp(vector<int>& nums, int target, int begin, int end)
{
if(begin > end) return end;
int mid = begin + (end - begin) / 2;
if(nums[mid] > target) return binarySearchUp(nums, target, begin, mid - 1);
else return binarySearchUp(nums, target, mid + 1, end);
}

};