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最新推荐文章于 2025-01-16 16:51:20 发布

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最新推荐文章于 2025-01-16 16:51:20 发布

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原文链接：http://www.cnblogs.com/my---world/p/10770861.html

本文详细解析了一个寻找连续子数组最大和的算法实现过程，通过动态规划思想，逐步展示了如何从给定数组中找到具有最大和的连续子数组，并附带了运行时间和每一步骤的详细解释。

package bao;

public class Solution {
     
    public static void main(String[] args) throws InterruptedException {
        int[] array = {1,-2,3,10,-4,7,2,-5};
        Long begintime = System.nanoTime();
        int result =  FindGreatestSumOfSubArray(array);
        Long endtime = System.nanoTime();
        System.out.println("连续子数组的最大和为："+result+",运行时间："+(endtime - begintime) + "ns");
        Thread.sleep(1000);
    }
 
    public static int FindGreatestSumOfSubArray(int[] array) throws InterruptedException {
           int len = array.length;
           if (len == 0){
            return 0;
           }
           int[] currentsum = new int[len];
           currentsum[0] = array[0];
           int greatsetsum = array[0];
           System.out.println("第1步：累加子数组和："+currentsum[0]+"，最大子数组和："+greatsetsum);
           Thread.sleep(1000);
           for(int i=1;i<array.length;i++){
                if(currentsum[i-1]>0){
                    currentsum[i] = currentsum[i-1] + array[i];
                }else{
                    currentsum[i] = array[i];
                }
                if(currentsum[i] > greatsetsum){
                    greatsetsum  = currentsum[i];
                }
               System.out.println("第"+(i+1)+"步：累加子数组和："+currentsum[i]+"，最大子数组和："+greatsetsum);
               Thread.sleep(1000);
           }
           return greatsetsum;
    }