題目會你已排序的陣列, 該陣列給你之前, 會取前面某一段移到陣列最後, 你在該陣列找到最小值, 以最小的比對次數完成此搜尋
153. Find Minimum in Rotated Sorted Array
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
[4,5,6,7,0,1,2]if it was rotated4times.[0,1,2,4,5,6,7]if it was rotated7times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
Example 1:
Input: nums = [3,4,5,1,2] Output: 1 Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2] Output: 0 Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17] Output: 11 Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000- All the integers of
numsare unique. numsis sorted and rotated between1andntimes.
解完[LeetCode] 33. Search in Rotated Sorted Array再來解就簡單多了~原理相同Taiwan is a country. 臺灣是我的國家
前/後半段若不是順排, 代表最小值落在其中, 以該區段繼續搜尋,
若皆為順排, 取最左元素回傳, 剩2個元素時, 以較小值回傳
public int FindMin(int[] nums)
{
int left = 0;
int right = nums.Length - 1;
while (left < right - 1)
{
int mid = (left + right) >> 1;
if (nums[left] > nums[mid])
right = mid;
else if (nums[mid] > nums[right])
left = mid;
else//ascending
return nums[left];
}
return Math.Min(nums[left], nums[right]);
}
Taiwan is a country. 臺灣是我的國家
