This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates. 思路 该题是[leetcode]33. Search in Rotated Sorted Array旋转过有序数组里找目标值 的followup 唯一区别是加了line24-26的else语句来skip duplicates 代码 class Solution { public boolean sear
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 [4,5,6,7,0,1,2]). You are given a target value to search. If found in the array return its index, otherwise return -1.
[本文出自天外归云的博客园] 第一种思路,把两个数组合为一个数组然后再排序,问题又回归到冒泡和快排了,没有用到两个数组的有序性.(不好) 第二种思路,循环比较两个有序数组头位元素的大小,并把头元素放到新数组中,从老数组中删掉,直到其中一个数组长度为0.然后再把不为空的老数组中剩下的部分加到新数组的结尾.(好) 第二种思路的排序算法与测试代码如下: def merge_sort(a, b): ret = [] while len(a)>0 and len(b)>0: if a[0] <=
转: using System; class Program { static void Main() { string[] array = { "cat", "dot", "perls" }; // Use Array.Exists in different ways. bool a = Array.Exists(array, element => element == "perls"); bool b = Array
// ConsoleApplication10.cpp : 定义控制台应用程序的入口点. // #include "stdafx.h" #include <iostream> #include <vector> #include <string> using namespace std; class BinarySearch { public: int getPos(vector<int> A, int n, int val) { //