[Leetcode] 34. Search for a Range 解题报告

题目：

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：

这还是典型的二分查找，关键在于怎么定义二分查找的边界条件。偷懒的做法是直接利用STL中提供的upper_bound和lower_bound方法，见代码片段1；不过估计这时签证官不会善罢甘休的，要求你写出bug free的upper_bound和lower_bound代码。那么请看代码片段2吧！

代码：

1、偷懒做法：

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> ret(2, -1);
auto it_start = lower_bound(nums.begin(), nums.end(), target);
auto it_end = upper_bound(nums.begin(), nums.end(), target);
if(it_start == it_end)
return ret;
ret[0] = distance(nums.begin(), it_start);
ret[1] = distance(nums.begin(), it_end) - 1;
return ret;
}
};

2、完整做法：

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> ret = {-1, -1};
if(nums.size() == 0)
return ret;
int lower = lower_bound(nums, 0, nums.size() - 1, target);
int upper =
4000
upper_bound(nums, lower, nums.size() - 1, target);  // to accelerate
if(lower == upper)
return ret;
ret[0] = lower;
ret[1] = upper - 1;
return ret;
}
private:
int lower_bound(vector<int>& nums, int left, int right, int target)
{
while(left <= right)
{
int mid = left + (right - left) / 2;
if(nums[mid] >= target)
right = mid - 1;
else
left = mid + 1;
}
return left;
}
int upper_bound(vector<int>& nums, int left, int right, int target)
{
while(left <= right)
{
int mid = left + (right - left) / 2;
if(nums[mid] > target)
right = mid - 1;
else
left = mid + 1;
}
return left;
}
};