Java 最长递增子序列的势O（n）解

我试图用递归（动态编程）来回答这个问题

从这篇文章中，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），我找不到它失败的案例。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));
    }

}

导入静态org.junit.Assert.assertEquals；
导入java.util.ArrayList；
导入java.util.array；
导入java.util.List；
导入org.junit.Test；
公共类LongestIncreasingSubseq{
公共静态void main（字符串[]args）{
int[]arr={0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15,1}；
getLongestSubSeq（arr）；
}
公共静态列表getLongestSubSeq（int[]arr）{
列表索引=longestrecurive（arr，0，arr.length-1）；
列表结果=新建ArrayList（）；
for（整数i：索引）{
结果。添加（arr[i]）；
}
System.out.println（result.toString（））；
返回结果；
}
私有静态列表longestrecurive（int[]arr，int start，int end）{
如果（开始==结束）{
List singleton=new ArrayList（）；
singleton.add（开始）；
返回单身；
}
List bestRightSubsequence=longestrecurive（arr，start+1，end）；//递归调用数组到下一个开始索引
if（bestRightSubsequence.size（）==1&&arr[start]>arr[bestRightSubsequence.get（0）]）{
bestRightSubsequence.set（0，开始）；//更大的结束允许更多的可能性
}else if（arr[start]1&&arr[start]

由于T（n）=T（n-1）+O（1）=>T（n）=O（n）的关系，成本严格为O（n）

有人能找到一个失败的案例，或者有任何bug吗？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));
    }

}

更新： 感谢大家指出我在以前的实现中的错误。下面的最终代码通过了它曾经失败的所有测试用例

其思想是列出（计算）所有可能的递增子序列（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memonization（使用哈希表）来避免重新计算已经计算的子序列——因此对于每个起始索引，我们只计算一次所有递增的子序列

然而，在这种情况下，我不确定如何正式推导出时间复杂性，如果有人能解释这一点，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));
    }

}

导入静态org.junit.Assert.assertEquals；
导入java.util.ArrayList；
导入java.util.array；
导入java.util.HashMap；
导入java.util.List；
导入java.util.Map；
导入org.junit.Test；
公共类longestincreatingsubsequence{
公共静态列表getLongestSubSeq（int[]arr）{
列表最长=新的ArrayList（）；
对于（int i=0；ilongestOnRight.size（））{
最长=最长或最短；
}
}
}
最长。添加（0，startIndex）；
List cloneOfLongest=新数组列表（最长）；
//必须始终缓存克隆，否则对象共享会把事情搞砸
cache.put（startIndex，cloneOfLongest）；//还记得这个子序列吗
返回时间最长；
}
@试验
公开无效测试（）{
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={0,0,0,0,0,1,1,1,1,2,3,0,0,0,1,1,1,0,0,3}；
int[]arr7={0,1,2,0,1,3}；
int[]
 @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));
    }

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

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))