算法说明:取中间位置的值与待查字比较.如果比待查字更大,则去列表的前半部分查找,如果比待查字小,则去列表的后半部分查找,直到找到这个待查字,或者返回没有找到这个待查字.其中给定的列表是从大到小排列的有序表 // 二分查找算法,待查的有序数组.待查的数字 public static int biSearch(int []array,int a) { int lo = 0; // 左边下标 int hi = array.length - 1; int mid; while(lo<=hi) { mid
//二分查找算法的实现 public static int binarySearch(int[] arr,int search) { int low=0; int high=arr.length-1; //找到中间的索引 int middleIndex=0; while (low<high) { middleIndex=(low+high+1)/2; if (arr[middleIndex]==search) { return middleIndex; } else if(arr[middleI
Java中常用的查找算法——顺序查找和二分查找 神话丿小王子的博客 一.顺序查找: a) 原理:顺序查找就是按顺序从头到尾依次往下查找,找到数据,则提前结束查找,找不到便一直查找下去,直到数据最后一位. b) 图例说明: 原始数据:int[] a={4,6,2,8,1,9,0,3}; 要查找数字:8 代码演示: import java.util.Scanner; /* * 顺序查找 */ public class SequelSearch { public static void main(St
up vote10down votefavorite I am trying to use Java 8 Streams to find elements in a LinkedList. I want to guarantee, however, that there is 1 and only 1 match to the filter criteria. Take this code: public static void main(String[] args) { LinkedList<
1.二分查找算法 package other; public class BinarySearch { /* * 循环实现二分查找算法arr 已排好序的数组x 需要查找的数-1 无法查到数据 */ public static int binarySearch(int[] arr, int x) { int low = 0; int high = arr.length-1; while(low <= high) { int middle = (low + high)/2; if(x == arr[
Java 8 – Filter a null value from a Stream package com.mkyong.java8; import java.util.List;import java.util.stream.Collectors;import java.util.stream.Stream; public class Java8Examples { public static void main(String[] args) { Stream<String> langua