Leetcode: Find Minimum in Rotated Sorted Array

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.

这道题跟Search in Rotated Sorted Array这道题很像，区别在于这道题不是找一个目标值，而是找最小值。所以与那道题的思路类似又有区别。类似在于同样是用binary search找目标区域，通过左边界和中间的大小关系来得到左边或者右边有序。区别在于这一次迭代不会存在找到target就终止的情况，iteration要走到 l > r 的情况才停止。所以这一次思路是确定中点哪一边有序之后，那一边的最小值就是最左边的元素。用这个元素尝试更新全局维护的最小值，然后迭代另一边。这样子每次可以截掉一半元素，所以最后复杂度等价于一个二分查找，是O(logn)，空间上只有四个变量维护二分和结果，所以是O(1)。

class Solution {
    public int findMin(int[] nums) {
        int res = Integer.MAX_VALUE;
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int m = (l + r) / 2;
            if (nums[m] >= nums[l]) {
                // rotation is happened in the latter half
                // [l, m] is in order
                res = Math.min(res, nums[l]);
                l = m + 1;
            }
            else {
                // rotation is happened in the first half
                // [m, r] is in order
                res = Math.min(res, nums[m]);
                r = m - 1;
            }
        }
        return res;
    }
}