Leetcode Find Minimum in Rotated Sorted Array 题解 Leetcode Find Minimum in Rotated Sorted Array 题目大意: 对一个有序数组翻转, 就是随机取前K个数,移动到数组的后面,然后让你找出最小的那个数,注意,K有可能是0,也就是没有翻转。 毫无疑问,遍历一次肯定可以找到,但这
publicintfindMin(int[]nums){intstart=0;intend=nums.length-1;while(start<end){intmid=(start+end)>>>1;if(nums[mid]>nums[end]){start=mid+1;}elseif(nums[mid]<nums[end]){end=mid;}else{//添加代码}}returnnums[start];} 可以想几个例子考虑下。 331233333^^midend上边的情况,mid==end,此...
思路:首先排除三种极端情况,空,只有一个元素,以及整个数组都是顺序排列的。 当顺序的数组随机旋转排列后,就分为两个顺序列入题目中的4567和012,寻找到中间数来和数组最后一个元素对比,如果大于的话说明最小的数在中间数的右边,如果小于的话说明最小数在中间数的左边,然 后继续按照二分法来找。注:不会出现中间数...
Leetcode Find Minimum in Rotated Sorted Array 题目大意: 对一个有序数组翻转, 就是随机取前K个数,移动到数组的后面,然后让你找出最小的那个数,注意,K有可能是0,也就是没有翻转。 毫无疑问,遍历一次肯定可以找到,但这样时间复杂度是O(n),如果你在面试的时候遇到这样的问题,你这样回答面试官肯定不会满意的...
int mid = (left+right)/2; if(a[mid] >= a[left]){ return helper(a, mid, right); }else{ return helper(a, left, mid); } } } https://chesterli0130.wordpress.com/2012/10/20/finding-the-minimum-in-a-sorted-rotated-array/
Swift – Find Minimum in Array To find the minimum element of an Array in Swift, callmin()method on this Array. Array.min() returns the minimum element by comparison. We can also provide a predicate to min() method, by which the comparison happens between elements. ...
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.
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...
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
I have a 7x7 array in which each row has a 0 element. I need to find the minimum value in each row that is not equal to zero. I've tried min(A(1,:)(A>0)) but it doesn't work. 댓글 수: 0 댓글을 달려면 로그인하십시...