153. Find Minimum in Rotated Sorted Array (M)

Find Minimum in Rotated Sorted Array (M)

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.

Example 1:

Input: [3,4,5,1,2] 
Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]
Output: 0

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题意

将一递增数列的随机后半部分与前半部分换位，得到新数组，在新数组中查找最小值(数组中没有重复的值)。

思路

目标是找到右半部分的第一个值，使用二分法实现。因为数组中不存在重复的值，所以根据nums[mid]与nums[0]大小关系可以很容易判断出mid落在前半部分还是后半部分，根据不同情况处理二分。

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代码实现

class Solution {
    public int findMin(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] >= nums[0]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        // left最终可能落在数组外，这说明数组本身就是一个递增数组。
        return left >= nums.length ? nums[0] : nums[left];
    }
}