动态规划----求某一数组中的一个连续段的元素和

法一：用链表来存数据，需要查找的时候，从头遍历取数据，最简单的思路，最麻烦的代码╮(╯_╰)╭

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Scanner;

//where  stands for dynamic programming
public class ThreeZeroThree{
    //Initialize an array
    public static int[] NumArray() {
    ArrayList<Integer> list = new ArrayList<Integer>();
    
     System.out.print("please input a value of n to stands for the size of the array");    
     Scanner s=new Scanner(System.in);
     int n=s.nextInt();
     for(int i=0;i<n;i++){
        Scanner count=new Scanner(System.in);
        int value= count.nextInt();
        list.add(value);
       }
      int length=list.size();
      int[] array = new int[length];
      for(int j=0;j<length;j++)
      array[j]=list.get(j);
      return array;
         
        
    }
    
public static void sumRange(int shuzi[],int a,int b){
    int sum=0;
    for(int k=a;k<=b;k++)
    sum=sum+shuzi[k];
    System.out.print(sum);
    
}
        
public static void main(String[] args){
    int a[]=NumArray();
     System.out.print("please input the value of The lower bound");    
     Scanner s=new Scanner(System.in);
     int n=s.nextInt();
     System.out.print("please input the value of The upper bound");    
     Scanner s2=new Scanner(System.in);
     int n2=s2.nextInt();
     sumRange(a,n,n2);
    
    }
    
    }

动态规划，先把和存起来，就不用每次重复遍历。

public class NumArray {

    public int[] sums;

    public NumArray(int[] nums) {
        if (nums == null) {
            sums = null;
        } else if (nums.length == 0) {
            sums = new int[0];
        } else {
            this.sums = new int[nums.length];
            sums[0] = nums[0];
            for (int i = 1; i < nums.length; i++) {
                sums[i] = sums[i - 1] + nums[i];
            }
        }
    }

    public int sumRange(int i, int j) {
        if (sums == null) {
            return 0;
        }
        if (i >= sums.length || j >= sums.length || i > j) {
            return 0;
        } else if (i == 0) {
            return sums[j];
        } else {
            return sums[j] - sums[i - 1];
        }
    }

}