Given an array of integers nums 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].
Example 1:
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| Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
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Example 2:
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| Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
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Constraints:
0 <= nums.length <= 10^5-10^9 <= nums[i] <= 10^9nums is a non decreasing array.-10^9 <= target <= 10^9
二分查找
题目等价于寻找第一个大于等于target的index left和最后一个小于等于target的index right
如果nums中没有target，则必有 left > right，返回[-1, -1]即可
如果有target，则返回[left, right]即为所求
而寻找left和right用广义的二分查找即可解决
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| class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
if(nums.size() == 0)
return vector<int>({-1,-1});
if(target < nums.front() || target > nums.back())
{
return vector<int>({-1,-1});
}
int left = getLeftIndex(nums,target);
int right = getRightIndex(nums,target);
if(left > right)
{
return vector<int>({-1,-1});
}
return vector<int>({left,right});
}
private:
int getLeftIndex(const vector<int>& nums, int target)
{
int i = 0;
int j = nums.size() - 1;
while(i <= j)
{
int mid = i + (j - i) / 2;
if(nums[mid] >= target)
{
j = mid - 1;
}
else
{
i = mid + 1;
}
}
return i;
}
int getRightIndex(const vector<int>& nums, int target)
{
int i = 0;
int j = nums.size() - 1;
while(i <= j)
{
int mid = i + (j - i) / 2;
if(nums[mid] <= target)
{
i = mid + 1;
}
else
{
j = mid - 1;
}
}
return j;
}
};
|