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[Leetcode]154. Find Minimum in Rotated Sorted Array II(C++)

题目描述

题目链接:154. Find Minimum in Rotated Sorted Array II

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

  • [4,5,6,7,0,1,4] if it was rotated 4 times.
  • [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

例子

例子 1

Input: nums = [1,3,5]
Output: 1

例子 2

Input: nums = [2,2,2,0,1]
Output: 0

Constraints

  • n == nums.length
  • 1 <= n <= 5000
  • `-5000 <= nums[i] <= 5000
  • nums is sorted and rotated between 1 and n times.

Follow up

This is the same as Find Minimum in Rotated Sorted Array but with duplicates. Would allow duplicates affect the run-time complexity? How and why?

解题思路

这道题是在 [Leetcode]153. Find Minimum in Rotated Sorted Array(C++) 的基础上添加了元素有可能会重复的情况。基本思路是上一题差不多,但是当 Left == Right && Mid == Left时,不能直接判断来确定 Mid 当前是在前半段还是后半段,这个时候我们只能通过右移左边界使得 Left != Right 才能继续判断,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) {
// move the left boundary if its equal to right boundary
while (left < right && nums[left] == nums[right]) {
left++;
}

// if the array is already sorted or there's only 1 element, break
if (left == right || nums[left] < nums[right]) {
break;
}

int mid = (left + right) / 2;
if (nums[mid] >= nums[left]) {
left = mid + 1;
} else {
right = mid;
}
}

return nums[left];
}
};
  • 时间复杂度: O(log n) -> O(n)
  • 空间复杂度: O(1)

GitHub 代码同步地址: 154.FindMinimumInRotatedSortedArrayIi.cpp

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