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。只有分为两种情况...
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. The array may contain duplicates. 这道题目就是Find Minimum in Rotated Sorted Array的增强版,在数组中增加了重复的元素。 具体的解法...
153. Find Minimum in Rotated Sorted Array 题目 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 min element. You may assume no duplicate exists in the array. Example ...
主要思路还是跟Search in Rotated Sorted Array差点儿相同。还是通过左边界和中间的大小关系来得到左边或者右边有序的信息。假设左半边有序。那么左半边最小就是左边第一个元素,能够和当前最小相比取小的,然后走向右半边。否则,那么就是右半半边第一个元素,然后走向左半边。这样子每次能够截掉一半元素,所以最后复杂...
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
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 of unique elements, return the minimum element of this array. ...
Search in rotated sorted array - Leetcode 33 - Python 呼吸的chou 0 0 Code-It-Yourself! Tetris - Programming from Scratch (Quick and Simple C++) 呼吸的chou 4 0 Decode Ways - Dynamic Programming - Leetcode 91 - Python 呼吸的chou 0 0 ...
还有一种是start+1 > end,也就是start == end,这种情况数组的长度是奇数,去任意一个返回即可 if(start==end)returnnums[start];if(nums[start]<nums[end])returnnums[start];elsereturnnums[end]; 精简下就是 if(nums[start] < nums[end])return nums[start];return nums[end]; ...
Find Minimum in Rotated Sorted Array II 理论上这题应该也用二分法,碰到起点终点相同的情况就只能进行遍历,但是那样代码会更复杂并且效率上并不见得提高多少,所以这里我直接采用遍历的方法。 实现代码: javapublic class Solution { public int findMin(int[] num) { ...
mid > start : 最小值在左半部分。 mid < start: 最小值在左半部分。 无论大于还是小于,最小值都在左半部分,所以 mid 和start 比较是不可取的。 mid 和end 比较 mid < end:最小值在左半部分。 mid > end:最小值在右半部分。 所以我们只需要把 mid 和end 比较,mid < end 丢弃右半部分(更新 en...