leetcode-algorithms-34 Find First and Last Position of Element in Sorted Array
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:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
解法
二分法查找
class Solution
{
public:
vector<int> searchRange(vector<int>& nums, int target)
{
vector<int> target_pos(2, -1);
int min = 0;
int max = nums.size() - 1;
int mid = 0;
int t = 0;
while(min <= max)
{
mid = (min + max) / 2;
t = nums[mid];
if (t == target)
{
target_pos.clear();
int pre = mid - 1;
bool have = false;
while(pre >= min && nums[pre] == target)
{
have = true;
--pre;
}
if (have) target_pos.push_back(++pre);
else target_pos.push_back(mid);
have = false;
pre = mid + 1;
while (pre <= max && nums[pre] == target)
{
have = true;
++pre;
}
if (have) target_pos.push_back(--pre);
else target_pos.push_back(mid);
break;
}
else if (t > target)
{
max = mid - 1;
}
else
{
min = mid + 1;
}
}
return target_pos;
}
};
时间复杂度: O(logn).
空间复杂度: O(1).