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. You may assume no duplicate exists in the array. SOLUTION 1: 这个题目trick的地方在于,它的旋转pivot次数未知。所以有可能它转...
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 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...
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. You may assume no duplicate exists in the array. 本题难度Medium。 二分查找法 【复杂度】...
但这样会有一个问题的,对于下边的例子,就会遇到死循环了。 问题出在,当数组剩两个的时候,mid = (start + end)/ 2,mid取的就是start。上图的例子,mid > end, 更新start = mid,start位置并不会变化。那么下一次mid的值也不会变,就死循环了。所以,我们要更新start = mid + 1,同时也使得start指向了最...
双指针,一头从左走,一头从右走任意一头出现,当前数小于左边且小于右边时,找到最小值,注意取余符号获取邻接的左边数和右边数min_num处理特殊情况没用二分法,反正是通过了,不管了, class Solution: def find…
My code: My test result: 这不是一个好方法。下面介绍 binary search 版方法。 My code: My test result: 一下子快...
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. ...
You may assume no duplicate exists in the array. class Solution { public: /** * @param num: a rotated sorted array * @return: the minimum number in the array */ int findMin(vector<int> &num) { // write your code here int l=0,r=num.size()-1; ...
Here, we are going to find the solution to find the minimum in rotated sorted array with C++ implementation.
Java program to find minimum element in a sorted and rotated array : If you want to practice data structure and algorithm programs, you can go through data structure and algorithm interview questions. In this post, we will see how to find minimum element in sorted and rotated array. Problem...