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    一维数组线性查找法

    public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int[] array = {11,23,234,12,3,45,21,233,23,32,123,321};
    System.out.println("请输入您要查找的数:");
    int num = input.nextInt();
    int index = -1;
    for (int i = 0; i < array.length; i++) {
    if(array[i]==num){
    index = i+1;
    break;
    }
    }
    if(index!=-1){
    System.out.println("find 在数组中的第"+index+"个位置");
    }else{
    System.out.println("not find!");
    }
    }

    一维数组线性查找法(求最大值和最小值)

    public static void main(String[] args) {
    int[] array = {11,23,234,12,3,45,21,233,23,32,123,321};
    int max = array[0];
    int min = array[0];
    for (int i = 0; i < array.length; i++) {
    if(array[i]>max){
    max = array[i];
    }
    if(array[i]<min){
    min = array[i];
    }
    }
    System.out.println("最大值:"+max);
    System.out.println("最小值:"+min);
    }

    二分查找法(数组的值必须是有序的)

    public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    System.out.println("请输入一个数:");
    int num = input.nextInt();
    int[] array = {1,2,3,4,5,6,7,8,9,0};
    int index = -1;
    int start = 0;
    int end = array.length-1;
    int middle;
    while(start<=end){
    //找到中间数的下标
    middle = (start+end)/2;
    if(num==array[middle]){
    index = middle;
    break;
    }
    if(num>array[middle]){
    start = middle+1;
    }
    if(num<array[middle]){
    end = middle-1;
    }
    }
    if(index!=-1){
    System.out.println("find 在数组中的第"+(index+1)+"个位置。");
    }else{
    System.out.println("not find!");
    }
    }

    冒泡排序法

    public static void main(String[] args) {
    int[] array = {11,23,234,12,3,45,21,233,23,32,123,321};
    for (int i = 0; i < array.length; i++) {
    for (int j = 0; j < array.length-1; j++) {
    if(array[j]>array[j+1]){
    int temp = array[j];
    array[j]=array[j+1];
    array[j+1]=temp;
    }
    }
    }
    for (int i = 0; i < array.length; i++) {
    System.out.print(array[i]+" ");
    }
    }
    冰冻三尺,并非一日之寒
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  • 原文地址:https://www.cnblogs.com/qjwcf/p/12856497.html
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