zoukankan      html  css  js  c++  java
  • 求数组小和(递归求解)


    import java.util.Arrays;

    /**
    * 求数组小和
    * 在一个数组中,一个数左边比它小的数的总和,叫数的小和,所有数的小和累加,叫数组小和。
    * 归并排序(递归)
    */
    public class SmallSum {

    public static void main(String[] args) {
    // 测试次数
    int times = 50000;

    int maxNum = 100;
    int maxSize = 100;
    for (int i = 0; i < times; i++) {
    // 生成随机数组
    int[] arr1 = generateArray(maxNum, maxSize);
    int[] arr2 = copyArray(arr1);
    // 小和
    int sum1 = smallSum(arr1);
    int sum2 = smallSum2(arr2);
    // 比较结果
    if (sum1 != sum2) {
    System.out.println("the result is error at " + i);
    System.out.println(sum1);
    System.out.println(sum2);
    System.out.println(Arrays.toString(arr1));
    System.out.println(Arrays.toString(arr2));
    return;
    }
    }
    System.out.println("the result is right");
    }

    public static int smallSum(int[] arr) {
    if (arr == null || arr.length < 2) {
    return 0;
    }
    return smallSum(arr, 0, arr.length - 1);
    }

    public static int smallSum(int[] arr, int left, int right) {
    if (left == right) {
    return 0;
    }
    int mid = left + ((right - left) >> 1);
    return smallSum(arr, left, mid)
    + smallSum(arr, mid + 1, right)
    + merge(arr, left, mid, right);
    }

    public static int merge(int[] arr, int left, int mid, int right) {
    int[] help = new int[right - left + 1];
    int res = 0;
    int index = 0;
    int po1 = left;
    int po2 = mid + 1;
    while (po1 <= mid && po2 <= right) {
    res += arr[po1] < arr[po2] ? (right - po2 + 1) * arr[po1] : 0;
    help[index++] = arr[po1] < arr[po2] ? arr[po1++] : arr[po2++];
    }
    while (po1 <= mid) {
    help[index++] = arr[po1++];
    }
    while (po2 <= right) {
    help[index++] = arr[po2++];
    }
    for (int i = 0; i < help.length; i++) {
    arr[left + i] = help[i];
    }
    return res;
    }

    /**
    * 生成随机数组
    *
    * @param maxNum 最大数
    * @param maxSize 数组最大大小
    * @return 随机数组
    */
    private static int[] generateArray(int maxNum, int maxSize) {
    int[] arr = new int[(int) (maxSize * Math.random())];
    for (int i = 0; i < arr.length; i++) {
    arr[i] = (int) (maxNum * Math.random());
    }
    return arr;
    }

    /**
    * 遍历求出数组小和
    *
    * @param arr 数组
    * @return 小和
    */
    public static int smallSum2(int[] arr) {
    int res = 0;
    for (int i = 1; i < arr.length; i++) {
    for (int j = 0; j < i; j++) {
    res += arr[j] < arr[i] ? arr[j] : 0;
    }
    }
    return res;
    }

    /**
    * 复制数组
    *
    * @param arr 要复制的数组
    * @return 复制的数组
    */
    private static int[] copyArray(int[] arr) {
    if (arr == null) {
    return null;
    }
    int[] copy = new int[arr.length];
    for (int i = 0; i < arr.length; i++) {
    copy[i] = arr[i];
    }
    return copy;
    }

    }

    /* 如有错误,欢迎批评指正 */
  • 相关阅读:
    App性能测试工具-PerfDog
    痛并快乐着
    SQLyog连接MySQL的前前后后
    组合模式
    Java并发编程:线程池的使用
    高效能人事的七个习惯
    Spring中Bean的生命周期及其扩展点
    (转)第一次有人把“分布式事务”讲的这么简单明了
    分布式事物
    mybatis学习笔记(2)基本原理
  • 原文地址:https://www.cnblogs.com/laydown/p/12813165.html
Copyright © 2011-2022 走看看