Java 数组中等于N的连续数的所有可能和

Java 数组中等于N的连续数的所有可能和,java,arrays,Java,Arrays,我试图从数组中的连续元素中找到所有可能的和,这些元素加起来就是一个特定的数字。 例如: 这是我的密码- int low = 0; int high = 0; int sum = a[0]; while(high < a.length) { if(sum < 13) { high++; if(high < a.length) {

我试图从数组中的连续元素中找到所有可能的和,这些元素加起来就是一个特定的数字。 例如:

这是我的密码-

 int low = 0;
        int high = 0;
        int sum = a[0];
        while(high < a.length) {
            if(sum < 13) {
                high++;

                if(high < a.length) {
                    sum+= a[high];

                }
            } else if(sum > 13) {
                sum-=a[low];
                low++;

            }
            if(sum == 13) {
                for(int i=low;i<=high;i++) {
                    System.out.println(a[i]);
                }
                System.out.println("new ");
                low++;
                high++;
                sum = 0;
                //return;
            }
        }
int-low=0;
int高=0;
整数和=a[0];
while(高13){
总和-=a[低];
低++;
}
如果(总和=13){

对于(int i=低;i在找到第一个序列后,没有正确重置变量:

        if(sum == 13) {
            for(int i=low;i<=high;i++) {
                System.out.println(a[i]);
            }
            System.out.println("new ");
            low++;
            high++; // change to high = low; // since you want your loop to test 
                    // sequences that start at the new (post incremented) low
            sum = 0; // change to sum = a[low]; // since the initial sum is the value of
                     // the first element in the new sequence
        }
if(总和=13){

对于(int i=low;i,如果数组可能包含零或负数

    int a[] = {4,7,2,1,3,8,5,-1,1};
    int length = a.length;
    for (int low = 0; low < length; ++low) {
        int sum = a[low];
        for (int high = low + 1; high < length; ++high) {
            sum += a[high];
            if (sum == 13) {
                for (int k = low; k <= high; ++k)
                    System.out.print(a[k] + " ");
                System.out.println();
            }
        }
    }

或者,您也可以保持该行的状态,并将
sum=0;
更改为
if(high
和顺便说一句。如果你想在数组中支持零,你只需要删除行
low++
;数组是否只包含正数?是的,如果我想支持-ve数字,我如何处理它们?
    int a[] = {4,7,2,1,3,8,5,-1,1};
    int length = a.length;
    for (int low = 0; low < length; ++low) {
        int sum = a[low];
        for (int high = low + 1; high < length; ++high) {
            sum += a[high];
            if (sum == 13) {
                for (int k = low; k <= high; ++k)
                    System.out.print(a[k] + " ");
                System.out.println();
            }
        }
    }
4 7 2 
7 2 1 3 
8 5 
8 5 -1 1