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 (num[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...
153. Find Minimum in Rotated Sorted Array Suppose a sorted array is rotated at some pivot unknown to you beforehand. 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. 注意是旋转...
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
You may assume no duplicate exists in the array. AI检测代码解析 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 ...
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 ...
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. ...
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]:...
57. Minimum Element in Sorted & Rotated Array Write a program in C to find the minimum element in a sorted and rotated array. Expected Output : The given array is : 3 4 5 6 7 9 2 The minimum element in the above array is: 2 ...
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 丢弃右半部分(更新 end = mid),mid > end 丢弃左半部分(...