intfindMinVal(intmin,intmax,constvector<int> &rotateArray) {//取三个点的值intminVal = rotateArray[min];//最左边的点的值intmaxVal = rotateArray[max];//最右边的点的值intmidVal = rotateArray[(min + max) /2];//中间的点的值//终止条件if((max - min) <=1)returnminVal > maxVal ?max...
与0153. Find Minimum in Rotated Sorted Array 相比,允许数组中存在重复的值,这带来的问题是,很难根据一个元素与最左侧元素的大小关系来判断它是落在左区间还是右区间,如数组 [10, 10, 1, 10, 10, 10]。 因此在 0153. Find Minimum in Rotated Sorted Array 解法的基础上做出改动:只有当nums[mid]与nu...
Follow up for “Find Minimum in Rotated Sorted Array”: What if duplicates are allowed? Would this affect the run-time complexity? How and why? 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 ...
This is a follow up problem toFind Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 题目大意 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] )。请找出其中最小的元...
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 ...
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 thatrotatingan 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 that may containduplicates, returnthe minimum element of this array. ...
接Find Minimum in Rotated Sorted Array, 假设有重复数字。 Solution 解法依然是二分搜索,不过多了一种情况要考虑。 classSolution(object):deffindMin(self,nums):""" :type nums: List[int] :rtype: int """L,R=0,len(nums)-1whileL<Randnums[L]>=nums[R]:M=(L+R)/2ifnums[M]>nums[R]:...
https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ 允许重复,也就意味着会有a[l]==a[r],以及a[mid]==a[r]的情况出现。后者比较好办,从坐标图中看出直接r=mid即可。 前者会有一个问题是当a[mid]==a[r]时,无法确定这个中间值在折线上的位置,此时退化为线性搜索,令l++。
Suppose a sorted array 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. The array may contain duplicates. 解法1:本题旋转数组最小值的扩展,这次数组中可以存在重复元素了。上题因数组中无重复元素,因此...