一、题目描述 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. 二、分析
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. The array may contain duplicates. 这道题目就是Find Minimum in Rotated Sorted Array的增强版,在数组中增加了重复的元素。 具体的解法...
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 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 ...
Find Minimum in Rotated Sorted Array II 理论上这题应该也用二分法,碰到起点终点相同的情况就只能进行遍历,但是那样代码会更复杂并且效率上并不见得提高多少,所以这里我直接采用遍历的方法。 实现代码: javapublic class Solution { public int findMin(int[] num) { int min = num[0], length = num.length...
详见:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/ Java实现: class Solution { public int findMin(int[] nums) { int low=0; int high=nums.length-1; int mid=0; while(nums[low]>=nums[high]){
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Find Minimum in Rotated Sorted Array. class Solution { public int findMin(int[] nums) { return findMin(nums, 0, nums.length - 1); } public int findMin(int[] nums, int left, int right) { ...
It's known that it's always possible to find both maximal and minimal elements of the array by using no more than comparisons, where⌈x⌉is the result of roundingxup. Write the program that will find positions of the minimum and the maximum in the jury's array of lengthn, by using...
2.1.1725 Part 1 Section 22.4.2.1, array (Array) 2.1.1726 Part 1 Section 22.4.2.5, clsid (Class ID) 2.1.1727 Part 1 Section 22.4.2.7, date (Date and Time) 2.1.1728 Part 1 Section 22.4.2.11, filetime (File Time) 2.1.1729 Part 1 Section 22.4.2.19, null (Null) 2.1....
2.1.1725 Part 1 Section 22.4.2.1, array (Array) 2.1.1726 Part 1 Section 22.4.2.5, clsid (Class ID) 2.1.1727 Part 1 Section 22.4.2.7, date (Date and Time) 2.1.1728 Part 1 Section 22.4.2.11, filetime (File Time) 2.1.1729 Part 1 Section 22.4.2.19, null (Null) 2.1.1730...
for n in A: if n <= smallest: smallest += 1 res += smallest - n else: smallest = n return res Loading...leetcode.com/problems/minimum-increment-to-make-array-unique/discuss/197687/JavaC%2B%2BPython-Straight-Forward 每次看到这几个人都会觉得,我不是在写博客,我只是博客的搬运工: ...