如何在不依赖 java 中的键值的情况下按顺序从树图中获取输出

How to get output from a treemap sequentially without depend upon the key value in java

我有编程问题。我有两个数组。 nums1={1,7,11} 和 nums2={2,4,6}。我必须找出第 k 次从每个数组中取 1 个数字的最小总和。在我的示例中,k=3。所以最小的和是{1+2=3},{1+4=5},{1+6=7}。数组始终按排序顺序排列。

我已经用树图解决了这个问题。

开始吧:

public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
     Map<Integer,int[]> value=new TreeMap<>();
     int[]arr_hold=new int[2];
     int []solve_arr=new int[k];
     List<int[]> solve = new ArrayList<>();
     ArrayList<Integer>add_sum= new ArrayList<>();
     int sum=0;
     for(int i=0;i<nums1.length;i++){
         for(int j=0;j<nums2.length;j++){
             sum=nums1[i]+nums2[j];
             arr_hold[0]=nums1[i];
             arr_hold[1]=nums2[j];
             value.put(sum,arr_hold);
         }
     }


     return solve;
 }

问题: 1.When 我打印 System.out.println(value); 输出:

{3=[I@75412c2f, 5=[I@75412c2f, 7=[I@75412c2f, 9=[I@75412c2f, 11=[I@75412c2f, 13=[I@75412c2f, 15=[I@75412c2f, 17=[I@75412c2f}

现在为什么数组 arr_hold 的值显示不正确?

当我使用 Treemap 时,它已经以排序的格式向我显示输出。因此,如果我从树图中获取第 1、2、3 个值,我的问题就会得到解决。但是这里的约束是

 value.get() method

搜索元素取决于键值。因此,如果我正在循环从树图中获取 3 个最小值,它会给我空输出,因为键不匹配。

 for(int k1=0;k1<k;k1++){
         System.out.println(value.get(k1));
     }

输出:空

     null

     null 

我该如何解决我的问题以我的方式?

这个怎么样:

import java.util.ArrayList;
import java.util.List;

public class KSmallestPairs {
    public static void main(String[] args) {
        System.out.println(kSmallestPairs(new int[]{1,7,11}, new int[]{2,4,6}, 3));
    }

    public static List<SumPair> kSmallestPairs(int[] leftInts, int[] rightInts, int k) {
        if (k < 1) {
            throw new IllegalArgumentException("k (=" + k + ") must higher than 0!");
        } else if (leftInts.length * rightInts.length < k) {
            throw new IllegalArgumentException("k (=" + k
                    + ") cannot be higher than the length of the cartesian product (="
                    + leftInts.length * rightInts.length + ")");
        }

        final List<SumPair> sumPairs = new ArrayList<>();
        int minLeftIndex = 0;
        int minRightIndex = 0;
        for (int leftIndex = 0, rightIndex = 0;
             leftIndex < leftInts.length
                     && rightIndex < rightInts.length
                     && sumPairs.size() < k; ) {            
            final int leftInt = leftInts[leftIndex];
            final int rightInt = rightInts[rightIndex];
            sumPairs.add(new SumPair(leftInt, rightInt));

            if(leftIndex + 1 < leftInts.length && rightIndex + 1 < rightInts.length) {
                final int nextLeftInt = leftInts[leftIndex + 1];
                final int nextRightInt = rightInts[rightIndex + 1];
                final int sumOfLeftIntAndNextRightInt = leftInt + nextRightInt;
                final int sumOfNextLeftIntAndRightInt = nextLeftInt + rightInt;
                if(sumOfLeftIntAndNextRightInt < sumOfNextLeftIntAndRightInt) {
                    rightIndex++;
                } else {
                    leftIndex++;
                }
            } else if(leftIndex + 1 < leftInts.length) {
                leftIndex++;
                rightIndex = minRightIndex;
                minLeftIndex++;
            } else if(rightIndex + 1 < rightInts.length) {
                leftIndex = minLeftIndex;
                rightIndex++;
                minRightIndex++;
            }
        }
        return sumPairs;
    }

    static class SumPair {
        private final int leftInt;
        private final int rightInt;

        public SumPair(int leftInt, int rightInt) {
            this.leftInt = leftInt;
            this.rightInt = rightInt;
        }

        public int getLeftInt() {
            return leftInt;
        }

        public int getRightInt() {
            return rightInt;
        }

        public int getSum() {
            return leftInt + rightInt;
        }

        @Override
        public String toString() {
            return leftInt + "+" + rightInt + "="+ getSum();
        }
    }
}

输出:

[1+2=3, 1+4=5, 1+6=7, 7+2=9]

它急切地找到 k 个最小的和对。