algorithm - 产生给定总和和乘积的子数组

Question

给定一个长度为 N 的数组，你将如何找到最小长度其和为 S 且其积为 P 的连续子数组。例如 5 6 1 4 6 2 9 7 for S = 17, Ans = [6, 2, 9] for P = 24, Ans = [4 6]。

Answer 1

就从左到右，对所有数字求和，如果总和 > S，则丢弃左边的数。

import java.util.Arrays;

public class test {
    public static void main (String[] args) {
        int[] array = {5, 6, 1, 4, 6, 2, 9, 7};
        int length = array.length;
        int S = 17;
        int sum = 0;                       // current sum of sub array, assume all positive
        int start = 0;                     // current start of sub array
        int minLength = array.length + 1;  // length of minimum sub array found
        int minStart = 0;                  // start of of minimum sub array found
        for (int index = 0; index < length; index++) {
          sum = sum + array[index];
          // Find by add to right
          if (sum == S && index - start + 1 < minLength) {
              minLength = index - start + 1;
              minStart = start;
          }
          while (sum >= S) {
            sum = sum - array[start];
            start++;
            // Find by minus from left
            if (sum == S && index - start + 1 < minLength) {
                minLength = index - start + 1;
                minStart = start;
            }
          }
        }
        // Found
        if (minLength != length + 1) {
            System.out.println(Arrays.toString(Arrays.copyOfRange(array, minStart, minStart + minLength)));
        }
    }
}

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对于你的例子，我认为是OR。

Product 与 sum 没有什么不同，除了计算。