Java publicclassSolution {/***@paramnum: a rotated sorted array *@return: the minimum number in the array*/publicintfindMin(int[] num) {if(num ==null|| num.length == 0)returnInteger.MIN_VALUE;intlb = 0, ub = num.length - 1;//case1: num[0] < num[num.length - 1]//if (...
https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array/ 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. 解题...
mid 和start 比较 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...
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 ...
Suppose a sorted array is rotated at some pivot unknown to you beforehand. 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. 思路: 有序数组旋转后,如果mid元素比low大,则左边有序,右边乱序,考虑最左元素是否最小元素...
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] )。
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. ...
minimum肯定在右半部分最下面那个数。我们可以先取array最右边的数,然后跟middle 比,如果比middle大证明我们在右边,如果比middle小证明我们在左边。【这里不能取最左边的一开始,因为也许array 并没有rotated,一条直线】 在确认了是在哪一边以后,就好办了。
Find Minimum in Rotated Sorted Array 其实直接遍历也是可以也可以AC,但是更加优化的解法应该是采用二分查找的思想。如果中间值比起点大就收缩起点,如果中间值比终点大就收缩终点,直到收缩到起点和终点相邻,也就是找到了翻转的部分。这里需要注意的是,数组可能是没有翻转过的,所以mid的初值赋0。