题目描述

题目链接：153. Find Minimum in Rotated Sorted Array

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

例子

例子 1

Input: nums = [3,4,5,1,2] Output: 1 Explanation: The original array was [1,2,3,4,5] rotated 3 times.

例子 2

Input: nums = [4,5,6,7,0,1,2] Output: 0 Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

例子 3

Input: nums = [11,13,15,17] Output: 11 Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

解题思路

这道题要求找到在一个旋转后的排序数组中找到最小值，直接遍历显然不是最优值。可以尝试使用二分查找，二分查找的关键是确定收缩左边界还是右边界。假设我们将数组分成两部分 A 作为前半部分，B 作为后半部分，显然 A 中所有值都比 B 大。在每一次二分查找中有以下几种情况：

• Left < Right：由于数组不包括重复数字，显然这种情况下数组本身有序，返回第一个元素即可
• Mid > Left: 可以确定 Mid 在 A 中，因此收缩左边界：Left = Mid + 1
• Mid <= Left：可以确定 Mid 在 B 中，因此收缩右边界，但是这里注意我们要获取的是 B 中第一个元素，所以 Mid 本身还有合理的搜索范围中，因此： Right = Mid

最后当 Left == Right,其所处位置就在 B 中第一个元素中，也就是我们的目标位置，返回该值即可，代码如下：

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0;
        int right = nums.size() - 1;
        while (left < right) {
            int mid = (left + right) / 2;
            if (nums[left] <= nums[right]) {
                break;
            }
            if (nums[mid] >= nums[left]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return nums[left];
    }
};

• 时间复杂度: O(log n)
• 空间复杂度: O(1)

GitHub 代码同步地址： 153.FindMinimumInRotatedSortedArray.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions