class Solution: # @param nums: The integer array # @param target: Target number to find # @return the first position of target in nums, position start from 0 def binarySearch(self, nums, target): if len(nums) == 0: return -1 start, end = 0, len(nums)
Find Minimum in Rotated Sorted Array Question Solution 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 ar
有一个无序序列[37,99,73,48,47,40,40,25,99,51],先进行排序打印输出,分别尝试插入20/40/41 数值到序列中合适的位置,保证其有序. 1.for 循环实现 第一种实现,利用嵌套for循环,每次迭代出来的数值进行比较.如果比原表中的数值小,则插入到这个数左面. lst1 = [37,99,73,48,47,40,40,25,99,51] lst = sorted(lst1) for x in (20,40,41,100): i = 0 for i,v
Find Minimum in Rotated Sorted Array II 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 be
Binary Search 二分法方法总结 code教你做人:二分法核心思想是把一个大的问题拆成若干个小问题,最重要的是去掉一半或者选择一半. 二分法模板: public int BinarySearchTemplate(int[] nums,int target) { if(nums == null || nums.length == 0) return -1; int lo = 0; int hi = nums.length - 1; //A: lo < hi [1,2]找1 找last p
二分法模板 非递归版本: public class Solution { /** * @param A an integer array sorted in ascending order * @param target an integer * @return an integer */ public int findPosition(int[] nums, int target) { if (nums == null || nums.length == 0) { return -1; } i
