153. Find Minimum in Rotated Sorted Array

最新推荐文章于 2021-09-07 22:51:22 发布

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最新推荐文章于 2021-09-07 22:51:22 发布

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CC 4.0 BY-SA版权

文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018154065

本文介绍了一种在旋转排序数组中查找最小元素的有效算法。通过二分搜索策略，文章详细阐述了如何在O(log n)的时间复杂度内找到最小元素，避免了线性搜索的高时间成本。

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

难度：medium

题目：假设一个按升序排序的数组在某个未知的轴上旋转。假定数组中没有重复元素。

思路：分情况二叉搜索

Runtime: 0 ms, faster than 100.00% of Java online submissions for Find Minimum in Rotated Sorted Array.
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Find Minimum in Rotated Sorted Array.

class Solution {
    public int findMin(int[] nums) {
        return findMin(nums, 0, nums.length - 1);
    }
    
    public int findMin(int[] nums, int left, int right) {
        int mid = left + (right - left) / 2;
        // 升序
        if (nums[left] <= nums[mid] && nums[mid] <= nums[right]) {
            return nums[left];
        }
        // 降序
        if (nums[left] >= nums[mid] && nums[mid] >= nums[right]) {
            return nums[right];
        }
        // 升序占大降序占小
        if (nums[left] >= nums[mid]) {
            return findMin(nums, left, mid);
        } 
        // 升序占小降序占大
        return findMin(nums, mid, right);
    }
}