source: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array
C/C++ Solution to LeetCode problem 153. Find Minimum in Rotated Sorted Array.
Problem
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 rotated4
times.[0,1,2,4,5,6,7]
if it was rotated7
times.
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.
Examples
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.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
- All the integers of
nums
are unique. nums
is sorted and rotated between1
andn
times.
Solution
The logic is pretty similar to the problem #33:
- We identify the sorted part of the array:
- If the value at the lower index
<
current min value:- Update minimum value.
- Next search will be in this part of the array.
- If not, then the minimum value is on the unsorted part.
- If the value at the lower index
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class Solution {
public:
int findMin(vector<int>& nums) {
int min = nums[0];
int l = 0;
int h = nums.size() - 1;
int m;
while (l <= h) {
m = (l + h) / 2;
if (nums[m] < min)
min = nums[m];
if (nums[l] <= nums[m]) {
if (nums[l] < min)
h = m - 1;
else
l = m + 1;
} else {
if (nums[l] < min)
l = m + 1;
else
h = m - 1;
}
}
return min;
}
};