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153. Find Minimum in Rotated Sorted Array
Medium
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.
Solution
public class Solution {
private int FindMinUtil(int[] nums, int l, int r) {
if (l == r) {
return nums[l];
}
int mid = (l + r) / 2;
if (mid == l && nums[mid] < nums[r]) {
return nums[l];
}
if (mid - 1 >= 0 && nums[mid - 1] > nums[mid]) {
return nums[mid];
}
if (nums[mid] < nums[l]) {
return FindMinUtil(nums, l, mid - 1);
} else if (nums[mid] > nums[r]) {
return FindMinUtil(nums, mid + 1, r);
}
return FindMinUtil(nums, l, mid - 1);
}
public int FindMin(int[] nums) {
int l = 0;
int r = nums.Length - 1;
return FindMinUtil(nums, l, r);
}
}