153. Find Minimum in Rotated Sorted Array

Description

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.
You may assume no duplicate exists in the array.

Solution

Binary-search, time O(logn), space O(1)

由于mid有可能跟left相等，但一定小于right，所以用nums[mid]和nums[right]的比较作为条件比较方便，不用考虑相等的情况。具体解释如下。

1. loop is left < right, which means inside the loop, left always < right
2. since we use round up for mid, and left < right from (1), right would never be the same as mid
3. Therefore, we compare mid with right, since they will never be the same from (2)
4. if nums[mid] < nums[right], we will know the minimum should be in the left part, so we are moving right.
   We can always make right = mid while we don’t have to worry the loop will not ends. Since from (2), we know right would never be the same as mid, making right = mid will assure the interval is shrinking.
5. if nums[mid] > nums[right], minimum should be in the right part, so we are moving left. Since nums[mid] > nums[right],mid can’t be the minimum, we can safely move left to mid + 1, which also assure the interval is shrinking

class Solution {
    public int findMin(int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (nums[mid] < nums[right]) {  // min could be min
                right = mid;
            } else {    // mid couldn't be min
                left = mid + 1;
            }
        }
        
        return nums[right];
    }
}