Leetcode | Find Minimum in Rotated Sorted Array I && II

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

You may assume no duplicate exists in the array.

二分查找就行了。

class Solution {
public:
    int findMin(vector<int> &arr) {
        if (arr.empty()) return -1;
        int l = 0, h = arr.size() - 1, m, min = INT_MAX;
        while (l <= h) {
            m = l + (h - l) / 2;
            if (arr[m] >= arr[l]) {
                if (min > arr[l]) min = arr[l];
                l = m + 1;
            } else {
                if (arr[m] < min) min = arr[m];
                h = m - 1;
            }
        }
        return min;
    }
};

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?
Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

class Solution {
public:
    int findMin(vector<int> &arr) {
        if (arr.empty()) return -1;
        int l = 0, h = arr.size() - 1, m, min = INT_MAX;
        while (l <= h) {
            m = l + (h - l) / 2;
            if (arr[m] > arr[l]) {
                if (min > arr[l]) min = arr[l];
                l = m + 1;
            } else if (arr[m] < arr[l]) {
                if (arr[m] < min) min = arr[m];
                h = m - 1;
            } else {
                if (arr[m] < min) min = arr[m];
                l++;
            }
        }
        return min;
    }
};