Suppose an array sorted in ascending order 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.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

This is a follow up problem to Find Minimum in Rotated Sorted Array.
Would allow duplicates affect the run-time complexity? How and why?

solution

time complexity : O(logn) (worst:O(n))
space complexity : O(1)

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

考慮陣列中 l, m, r 元素之間的關係。
在有rotated的情況下，
若 m > r 代表min在 m+1~r 之間，例：345678012
若 l > m 代表min在 l+1~m 之間，例：780123456
上述兩者都不是的話，也不代表就是 l < m < r ，最差情況像是3133333，無法直接輸出最左邊的元素。
就只能一一從r縮減搜索範圍。