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[Leetcode]153. Find Minimum in Rotated Sorted Array(C++)

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 MidA 中,因此收缩左边界:Left = Mid + 1
  • Mid <= Left:可以确定 MidB 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:

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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