Question

我试图从数组中连续元素中找到所有可能的总和，这些元素会添加到特定数字。例如： -

a[] = {4,7,2,1,3,8,5}; and N = 13,
Output = {4,7,2},{7,2,1,3},{8,5}

这是我的代码 -

 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;
            }
        }

但输出只返回一组{4,7,2}。我无法打印其他套装。任何人都可以帮我解决这个问题

Answer 1

找到第一个序列后，您没有正确重置变量：

        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
        }

Answer 2

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();
            }
        }
    }

结果：

连续数字的所有可能总和，等于数组中的N.

2 个答案: