java - 我们如何获得左右子数组总和相同的数组的索引‽

Question

我已完成以下编程练习:Equal Side of an Array 。声明如下:

You are going to be given an array of integers. Your job is to take that array and find an index N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that would make this happen, return -1.

For example:

Let's say you are given the array {1,2,3,4,3,2,1}: Your function will return the index 3, because at the 3rd position of the array, the sum of left side of the index ({1,2,3}) and the sum of the right side of the index ({3,2,1}) both equal 6.

Let's look at another one. You are given the array {1,100,50,-51,1,1}: Your function will return the index 1, because at the 1st position of the array, the sum of left side of the index ({1}) and the sum of the right side of the index ({50,-51,1,1}) both equal 1.

Last one: You are given the array {20,10,-80,10,10,15,35} At index 0 the left side is {} The right side is {10,-80,10,10,15,35} They both are equal to 0 when added. (Empty arrays are equal to 0 in this problem) Index 0 is the place where the left side and right side are equal.

Note: Please remember that in most programming/scripting languages the index of an array starts at 0.

Input: An integer array of length 0 < arr < 1000. The numbers in the array can be any integer positive or negative.

Output: The lowest index N where the side to the left of N is equal to the side to the right of N. If you do not find an index that fits these rules, then you will return -1.

Note: If you are given an array with multiple answers, return the lowest correct index.

我已阅读用户 JensPiegsa 提供的以下答案。 Here you have the link to it.

import java.util.stream.IntStream;

public class Kata {
  public static int findEvenIndex(int[] arr) {
    return IntStream.range(0, arr.length)
        .filter(n -> IntStream.of(arr).limit(n).sum() == IntStream.of(arr).skip(n + 1).sum())
        .findFirst().orElse(-1);
  }
}

我想知道是否有一种方法可以代替循环遍历左右子数组相等的 Instream 过滤，然后返回第一个；当我们返回第一个相等时中断执行。

我想知道当我们只获取左右相等的子数组而不对其进行循环时，功能解决方案会是什么样子。

对于循环，我认为可能是:

public class Kata {
  public static int findEvenIndex(int[] arr) {
    int left = 0, right = 0;

    for(int i = 0; i < arr.length; i++, left = 0, right = 0){
      for(int j = 0; j < i; j++){
        left += arr[j];
      }
      for(int k = arr.length - 1; k > i; k--){
        right += arr[k];
      }
      if(left == right) return i;  
    }
    return -1;
  }
}

如何用一个功能性句子来完成它？

我还读过: Is there a subarray that sums to a target? Find the index of the subarray whose sum is minimum Make sums of left and right sides of array equal by removing subarray

Answer 1

如果你想要一个非 O(n^2) 的解决方案:

    public Integer getfirstIndexEqual(Integer [] nums){
        Integer sum = 0;
        for (int i = 0; i < nums.length; i++){
            sum += nums[i];
        }
        Integer half = 0;
        for (int i=0; i< nums.length; i++){
            if (half.floatValue() == (sum-nums[i]) / 2){
                return i;
            }
            half += nums[i];
        }
        return -1;
    }