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
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...
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。
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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&&...
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...