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));
}
}
成本严格为O（n），因为关系T（n）= T（n-1）+ O（1）=> T（n）= O（n）
任何人都可以找到这个失败的案例，或者有任何错误吗？非常感谢。
更新：感谢大家在以前的实施中指出我的错误。下面的最终代码传递了它曾经失败的所有测试用例。
我们的想法是列出（计算）所有可能的增加子序列（每个子序列从索引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）来避免重新计算已经计算的子序列 - 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时间复杂性 - 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));
}
}
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};
你的结果[0, 1, 3]不应该是[0,1,2,3]
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));
}
并且它失败了，因为它根据你的评论给出了输出{1,3,8}。
很抱歉成为坏消息的承担者，但这实际上是O（n2）。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] <arr [i]恰好出现总和（i-1,1，n）= n *（n-1）/ 2次，这肯定是O（n2）。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;其中第一个将触发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] <arr [i]，因为在递归开始展开之前没有任何内容放入缓存。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。
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++)。在这个循环中你叫longestSubseqStartsWith(arr, i);。看看longestSubseqStartWith实现我们看到for (int i = startIndex + 1; i < arr.length; i++)
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))