一、题目描述 Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.,0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 二、分析
Can you solve this real interview question? 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
所以就随便搜了下Garnker的,知道是Search in Rotated Sorted Array,那题是找目标值。这里是找最小值。 我是这样想的,分为几部分, 1.如果左边的值等于中间的值了,那么就剩下两个数要判断了,而且是相邻的。 2.如果中间的值大于左边的值,那么最小值肯定不是在最左边就只在右边一块了。 3.如果中间值小于左...
而Find Minimum in Rotated Sorted Array II这道题这考虑了元素反复的情况,这里仅仅是为了算法更加一般性。就把那个情况也包括进来,有兴趣的朋友能够看看Find Minimum in Rotated Sorted Array II对反复元素的分析哈。
This is a follow up problem to Find Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 描述 假设按照升序排序的数组在预先未知的某个点上进行了旋转。 ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
Leetcode Find Minimum in Rotated Sorted Array 题目大意: 对一个有序数组翻转, 就是随机取前K个数,移动到数组的后面,然后让你找出最小的那个数,注意,K有可能是0,也就是没有翻转。 毫无疑问,遍历一次肯定可以找到,但这样时间复杂度是O(n),如果你在面试的时候遇到这样的问题,你这样回答面试官肯...
在排序数组中查找元素的第一个和最后一个位置 题目描述: 给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。 你...
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. ...
0152-maximum-product-subarray.rs 0153-find-minimum-in-rotated-sorted-array.rs 0155-min-stack.rs 0167-two-sum-ii-input-array-is-sorted.rs 0169-majority-element.rs 0179-largest-number.rs 0190-reverse-bits.rs 0191-number-of-1-bits.rs 0198-house-robber.rs 0199-binary-tree-right-side-view....
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) { ...