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 这题难度有,因为他默认的是数组中所有的元素是不同的。只...
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] )。
Find the minimum element. You may assume no duplicate exists in the array. 思路分析:这题主要考察二分查找,在Rotated Sorted Array中找最小和Search in Rotated Sorted Array相似,仅仅只是此时须要不断把最左边的元素A[l]与A[m]比較 1. 假设A[m] < A[l] 那么应该在左边找,更新右边的边界right。
在这道题目里面,其实和33题LeetCode33. Search in Rotated Sorted Array很像,就是多了一步对-1的判定。 答案解析 二分法 intbinarySearch(int*nums,intnumsSize,inttarget,bool lower){intleft=0,right=numsSize-1,ans=numsSize;while(left<=right){intmid=(left+right)/2;if(nums[mid]>target||(lower&&...
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]. ...
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) { ...
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