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...
*@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 (num[lb] < num[ub]) return num[lb];// ...
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】 时间O(N) 空间 O(N) 【思路】 题目假设:You may assume no duplicate exists in the array. 这道题与其他Rotated Sorted Array题一样可以用改良的二分查找法来做。 原则是只要一直...
链接: https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array/ 旋转数组 查找最小值 最小值肯定在旋转轴处 ,所以 用二分搜索找出旋转轴就可以了 每次取中点,前后肯定有一截数是有序,另一截是无序..取无序的那部分继续搜索即可 class Solution { public: int findMin(vector<int> &num...
class Solution: def findMin(self, nums: List[int]) -> int: if len(nums) == 1: return nums[0] i = 0 nums_len = len(nums) j = nums_len - 1 min_num = nums[0] while (i <= j): r1 = nums[(i + 1) % nums_len] l1 = nums[(i - 1 + nums_len) % nums_len] c1 ...
* @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; while(l<r-1){// int mid=l+(r-l)/2; if(num[mid]<num[r]){ ...
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 arraynumsofuniqueelements, returnthe minimum element of this array. ...
* @param num: a rotated sorted array * @return: the minimum number in the array */public intfindMin(int[]num){// write your code hereint length=num.length;int min=num[length-1];for(int i=0;i<length;i++){if(num[i]<min)min=num[i];}returnmin;}}...
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. ...
println("Minimum element in the array : "+findMinimumElementRotatedSortedArray(arr,0,arr.length-1,5)); } public static int findMinimumElementRotatedSortedArray(int[] arr,int low,int high,int number) { int mid; while(low<high) { mid=low + ((high - low) / 2); if(arr[mid] > arr...