对最长增长子序列的潜在O（n）解

Question

我试图回答这个问题，只使用递归（动态编程） http://en.wikipedia.org/wiki/Longest_increasing_subsequence

从文章和SO周围，我意识到最有效的现有解决方案是O（nlgn）。我的解决方案是O（N），我找不到它失败的情况。我包括我使用的单元测试用例。

import static org.junit.Assert.assertEquals;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

import org.junit.Test;

public class LongestIncreasingSubseq {

    public static void main(String[] args) {
        int[] arr = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15, 1};
        getLongestSubSeq(arr);
    }

    public static List<Integer> getLongestSubSeq(int[] arr) {
        List<Integer> indices = longestRecursive(arr, 0, arr.length-1);
        List<Integer> result = new ArrayList<>();
        for (Integer i : indices) {
            result.add(arr[i]);
        }

        System.out.println(result.toString());
        return result;
    }

    private static List<Integer> longestRecursive(int[] arr, int start, int end) {
        if (start == end) {
            List<Integer> singleton = new ArrayList<>();
            singleton.add(start);
            return singleton;
        }

        List<Integer> bestRightSubsequence = longestRecursive(arr, start+1, end); //recursive call down the array to the next start index
        if (bestRightSubsequence.size() == 1 && arr[start] > arr[bestRightSubsequence.get(0)]) {
            bestRightSubsequence.set(0, start); //larger end allows more possibilities ahead
        } else if (arr[start] < arr[bestRightSubsequence.get(0)]) {
            bestRightSubsequence.add(0, start); //add to head
        } else if (bestRightSubsequence.size() > 1 && arr[start] < arr[bestRightSubsequence.get(1)]) {
            //larger than head, but still smaller than 2nd, so replace to allow more possibilities ahead
            bestRightSubsequence.set(0, start); 
        }

        return bestRightSubsequence;
    }

    @Test
    public void test() {
        int[] arr1 = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15, 1};
        int[] arr2 = {7, 0, 9, 2, 8, 4, 1};
        int[] arr3 = {9, 11, 2, 13, 7, 15};
        int[] arr4 = {10, 22, 9, 33, 21, 50, 41, 60, 80};
        int[] arr5 = {1, 2, 9, 4, 7, 3, 11, 8, 14, 6};
        assertEquals(getLongestSubSeq(arr1), Arrays.asList(0, 4, 6, 9, 11, 15));
        assertEquals(getLongestSubSeq(arr2), Arrays.asList(0, 2, 8));
        assertEquals(getLongestSubSeq(arr3), Arrays.asList(9, 11, 13, 15));
        assertEquals(getLongestSubSeq(arr4), Arrays.asList(10, 22, 33, 50, 60, 80));
        assertEquals(getLongestSubSeq(arr5), Arrays.asList(1, 2, 4, 7, 11, 14));
    }

}

由于T（n）= T（n-1）+ O（1）=＆gt;的关系，成本严格为O（n）。 T（n）= O（n）

任何人都可以找到失败的案例，或者有任何错误吗？非常感谢。

更新 谢谢大家指出我之前实施中的错误。下面的最终代码传递了它曾经失败的所有测试用例。

想法是列出（计算）所有可能的增加子序列（每个从索引i开始从0到N.length-1）并选择最长的子序列。我使用memoization（使用哈希表）来避免重新计算已计算的子序列 - 因此对于每个起始索引，我们只计算所有增加的子序列一次。

但是，我不确定如何在这种情况下正式推导出时间复杂度 - 如果有人能够阐明这一点，我将不胜感激。非常感谢。

import static org.junit.Assert.assertEquals;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import org.junit.Test;

public class LongestIncreasingSubsequence {

    public static List<Integer> getLongestSubSeq(int[] arr) {
        List<Integer> longest = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            List<Integer> candidate = longestSubseqStartsWith(arr, i);
            if (longest.size() < candidate.size()) {
                longest = candidate;
            }
        }

        List<Integer> result = new ArrayList<>();
        for (Integer i : longest) {
            result.add(arr[i]);
        }

        System.out.println(result.toString());
        cache = new HashMap<>(); //new cache otherwise collision in next use - because object is static
        return result;
    }

    private static Map<Integer, List<Integer>> cache = new HashMap<>();
    private static List<Integer> longestSubseqStartsWith(int[] arr, int startIndex) {
        if (cache.containsKey(startIndex)) { //check if already computed
            //must always return a clone otherwise object sharing messes things up
            return new ArrayList<>(cache.get(startIndex)); 
        }

        if (startIndex == arr.length-1) {
            List<Integer> singleton = new ArrayList<>();
            singleton.add(startIndex);
            return singleton;
        }

        List<Integer> longest = new ArrayList<>();
        for (int i = startIndex + 1; i < arr.length; i++) {
            if (arr[startIndex] < arr[i]) {
                List<Integer> longestOnRight = longestSubseqStartsWith(arr, i);
                if (longestOnRight.size() > longest.size()) {
                    longest = longestOnRight;
                }
            }
        }

        longest.add(0, startIndex);
        List<Integer> cloneOfLongest = new ArrayList<>(longest);
        //must always cache a clone otherwise object sharing messes things up
        cache.put(startIndex, cloneOfLongest); //remember this subsequence
        return longest;
    }

    @Test
    public void test() {
        int[] arr1 = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15, 1};
        int[] arr2 = {7, 0, 9, 2, 8, 4, 1};
        int[] arr3 = {9, 11, 2, 13, 7, 15};
        int[] arr4 = {10, 22, 9, 33, 21, 50, 41, 60, 80};
        int[] arr5 = {1, 2, 9, 4, 7, 3, 11, 8, 14, 6};
        int[] arr6 = {0,0,0,0,0,0,1,1,1,1,2,3,0,0,0,1,1,0,1,1,0,1,0,3};
        int[] arr7 = {0,1,2,0,1,3};
        int[] arr8 = {0,1,2,3,4,5,1,3,8};
        assertEquals(getLongestSubSeq(arr1), Arrays.asList(0, 4, 6, 9, 13, 15));
        assertEquals(getLongestSubSeq(arr2), Arrays.asList(0, 2, 8));
        assertEquals(getLongestSubSeq(arr3), Arrays.asList(9, 11, 13, 15));
        assertEquals(getLongestSubSeq(arr4), Arrays.asList(10, 22, 33, 50, 60, 80));
        assertEquals(getLongestSubSeq(arr5), Arrays.asList(1, 2, 4, 7, 11, 14));
        assertEquals(getLongestSubSeq(arr6), Arrays.asList(0,1,2,3));
        assertEquals(getLongestSubSeq(arr7), Arrays.asList(0,1,2,3));
        assertEquals(getLongestSubSeq(arr8), Arrays.asList(0, 1, 2, 3, 4, 5, 8));
    }

    public static void main(String[] args) {
        int[] arr1 = {7, 0, 9, 2, 8, 4, 1};
        System.out.println(getLongestSubSeq(arr1));
    }

}

Answer 1

您的程序在此测试用例中失败

int[] arr5 = {0,0,0,0,0,0,1,1,1,1,2,3,0,0,0,1,1,0,1,1,0,1,0,3};

您的结果[0, 1, 3] 不应该是[0,1,2,3]

Answer 2

刚才我使用以下测试用例尝试了你的算法：

 @Test
    public void test() {

      int[] arr1 = {0,1,2,3,4,5,1,3,8};
      assertEquals(getLongestSubSeq(arr1), Arrays.asList(0, 1, 2, 3, 4, 5, 8));
    }

它失败了，因为它给出了输出{1,3,8} 根据您的评论编辑。

Answer 3

很抱歉成为坏消息的承担者，但这实际上是O（n ² ）。我不确定你是否有更正式的东西，但这是我的分析：

consider the case when the input is sorted in descending order
  (longestRecursive is never executed recursively, and the cache has no effect)

getLongestSubSeq iterates over the entire input -> 1:n
  each iteration calls longestRecursive
  longestRecursive compares arr[startIndex] < arr[i] for startIndex+1:n -> i - 1

因此，比较arr [startIndex]＆lt; arr [i]恰好出现总和（i - 1,1，n）= n *（n - 1）/ 2次，这肯定是O（n ² ）。您可以通过发送按升序排序的输入来强制使用最大缓存。在这种情况下，getLongestSubSeq会调用longestRecursive n次;第一个将触发n - 1个递归调用，每个调用都会导致缓存未命中并运行i - 1比较arr [startIndex]＆lt; arr [i]因为在递归开始展开之前没有任何内容被放入缓存中。比较次数与我们绕过缓存的示例完全相同。事实上，比较的数量总是相同的;在输入中引入反转只会导致代码交换递归以进行迭代。

Answer 4

这是一个O（n ^ 2）算法。因为有两个循环。第二个循环隐藏在方法调用中。

这是第一个循环：for (int i = 0; i < arr.length; i++)。在这个循环中，你调用了longestSubseqStartsWith(arr, i);。查看longestSubseqStartWith实现，我们看到for (int i = startIndex + 1; i < arr.length; i++)

Answer 5

这是我在python3.x中的潜在O（N）解决方案：

l = list(map(int,input().split()))
t = []
t2 = []
m = 0
for i in l:
    if(len(t)!=0):
        if(t[-1]<=i):
            if(t[-1]!=1):
                 t.append(i)
        else:
            if(len(t)>m):
                t2 = t
                m = len(t)
            t = [i]
    else:
        t.append(i)
print(t2,len(t2))