[LeetCode]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]

.

[LeetCode Source]

思路：

1）先用二分查找找到点，然后再左右扩展，找到范围。如果有m个要搜寻点，算法复杂度为O(lgN+m),m<=N。最好的情况下是O(lgN)，只有一个点的时候；最差情况O(N)，全是要搜寻点的时候。这种做法可以Ac。

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> result(2,-1);
int left = 0;
int right = nums.size()-1;
int mid = 0;
int first = -1;
int second = -1;
while(left<=right){
mid = left+(right-left)/2;
if(nums[mid] > target)
right = mid - 1;
if(nums[mid] < target)
left = mid + 1;
if(nums[mid] == target){
first = mid;
second = mid;
break;
}
}
for(int i=mid-1;i>=0;i--){
if(nums[i]==target)
first--;
else break;
}
for(int i=mid+1;i<nums.size();++i){
if(nums[i]==target)
second++;
else break;
}
result[0] = first;
result[1] = second;
return result;
}
};

2）可能是最优解法。两次二分查找，一次是找到左边界点，第二次找到右边界点。

找左边界点时，修改二分查找的规则，如果找到点，还要继续往左二分直到找到左边界点；

找右边界点时，同理。

特别注意break的处理，当明显找不到点时break掉循环，对于找左边界点就是nums[right]<target。

这种解法复杂度是O(lgN)

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int hit = -1;
int left=0, right=nums.size() - 1;
int first = 0, second = 0;

vector<int> ret(2, -1);

while(left <= right)
{
int mid = left + (right - left) / 2;
if(nums[right] < target)    break;

if(nums[mid] >= target)
{
if(nums[mid] == target) hit = mid;
right = mid - 1;
}
else
{
left = mid + 1;
}
}
if(hit == -1)   return ret;
first = hit;

left = hit;
right = nums.size() - 1;
hit = -1;

while(left <= right)
{
int mid = left + (right - left) / 2;
if(nums[left] > target)    break;

if(nums[mid] <= target)
{
if(nums[mid] == target) hit = mid;
left = mid + 1;
}
else
{
right = mid - 1;
}
}
if(hit == -1)   return ret;
second = hit;

ret[0] = first;
ret[1] = second;

return ret;
}
};