LeetCode - Medium - 1143. Longest Common Subsequence

心已赠人 2022-10-22 10:47 187阅读 0赞

## Topic ##

*  Dynamic Programming

## Description ##

[https://leetcode.com/problems/longest-common-subsequence/][https_leetcode.com_problems_longest-common-subsequence]

Given two strings `text1` and `text2`, return the length of their longest **common subsequence**. If there is no **common subsequence**, return `0`.

A **subsequence** of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

For example, `"ace"` is a subsequence of `"abcde"`.  
A **common subsequence** of two strings is a subsequence that is common to both strings.

**Example 1**:

Input: text1 = "abcde", text2 = "ace" 
    Output: 3  
    Explanation: The longest common subsequence is "ace" and its length is 3.

**Example 2**:

Input: text1 = "abc", text2 = "abc"
    Output: 3
    Explanation: The longest common subsequence is "abc" and its length is 3.

**Example 3**:

Input: text1 = "abc", text2 = "def"
    Output: 0
    Explanation: There is no such common subsequence, so the result is 0.

Constraints:

*  `1 <= text1.length, text2.length <= 1000`
 *  `text1` and `text2` consist of only lowercase English characters.

## Analysis ##

### 方法一：标准DP ###

动态规划五部曲:

1.  确定dp数组（dp table）以及下标的含义；
2.  确定递推公式；
3.  dp数组如何初始化；
4.  确定遍历顺序；
5.  举例推导dp数组。

**DP数组意，归纳递推式，初始元素值，定序来遍历，最后举个例**。

用动规五部曲分析如下

**1.确定dp数组（dp table）以及下标的含义**

dp\[i\]\[j\]表示 从text1的下标0开始截取长度为i的字符串 与 从text2的下标0开始截取长度为j的字符串 的最长公共子序列的长度。

也可以换句话说。

dp\[i\]\[j\]表示 从text1的下标0开始截取到下标i-1截取的字符串（包含下标i-1处的字符） 与 从text2的下标0开始截取到下标j-1截取的字符串（包含下标j-1处的字符） 的最长公共子序列的长度。

**2.确定递推公式**

主要就是两大情况：text1\[i - 1\] 与 text2\[j - 1\]相同，text1\[i - 1\] 与 text2\[j - 1\]不相同

1.  如果text1\[i - 1\] 与 text2\[j - 1\]相同，那么找到了一个公共元素，所以`dp[i][j] = dp[i - 1][j - 1] + 1;`。
2.  如果text1\[i - 1\] 与 text2\[j - 1\]不相同，那就看看text1\[0, i - 2\]与text2\[0, j - 1\]的最长公共子序列 和 text1\[0, i - 1\]与text2\[0, j - 2\]的最长公共子序列，取最大的，即：`dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);`。

代码如下：

if (text1[i - 1] == text2[j - 1]) { 
        dp[i][j] = dp[i - 1][j - 1] + 1;
    } else { 
        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
    }

**3.dp数组如何初始化**

先看看dp\[i\]\[0\]应该是多少呢？

test1\[0, i-1\]和空字符串的最长公共子序列自然是0，所以dp\[i\]\[0\] = 0;

同理dp\[0\]\[j\]也是0。

其他下标都是随着递推公式逐步覆盖，初始值是默认的即可。

总之，全部都是0，即默认的。

int[][] dp = new int[text1.length() + 1][text2.length() + 1];

**4.确定遍历顺序**

从递推公式，可以看出，有三个方向可以推出dp\[i\]\[j\]，如图：

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3UwMTE4NjMwMjQ_size_16_color_FFFFFF_t_70_pic_center]

那么为了在递推的过程中，这三个方向都是经过计算的数值，所以要**从前向后，从上到下**来遍历这个矩阵。

**5.举例推导dp数组**

以输入：text1 = “abcde”, text2 = “ace” 为例，dp状态如图：

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3UwMTE4NjMwMjQ_size_16_color_FFFFFF_t_70_pic_center 1]

最后红框dp\[text1.length()\]\[text2.length()\]为最终结果。

**再次重申**，dp\[i\]\[j\]表示 从text1的下标0开始截取长度为i的字符串 与 从text2的下标0开始截取长度为j的字符串 的最长公共子序列的长度。

### 方法二：方法一的空间优化 ###

进一步地观察，方法一的代码遍历时只需前一行以及目前一行的信息即可。因此，dp数组初始化**两行的**。

注意，用k ^ 1、k ^= 1来切换 dp\[0\] (第一行) 与 dp\[1\] (第二行)。

注意，m % 2 与 m & 1 意同。

### 方法三：方法一的空间优化Plus ###

再进一步地观察，方法一的代码遍历时只需前一行以及目前一行的前一元素信息即可。因此，dp数组**初始化一行**的，另加3个变量进行辅助。

最后，优化后的时空杂度分别为：O(m \* n)，O(min(m, n))。

### 参考资料 ###

1.  [动态规划：最长公共子序列][Link 1]
2.  [\[Java/Python 3\] Two DP codes of O(mn) & O(min(m, n)) spaces w/ picture and analysis][Java_Python 3_ Two DP codes of O_mn_ _ O_min_m_ n_ spaces w_ picture and analysis]

## Submission ##

public class LongestCommonSubsequence { 
    	// 方法一：标准DP
    	public int longestCommonSubsequence(String text1, String text2) { 
    
    		int[][] dp = new int[text1.length() + 1][text2.length() + 1];
    
    		for (int i = 1; i <= text1.length(); i++) { 
    			for (int j = 1; j <= text2.length(); j++) { 
    				if (text1.charAt(i - 1) == text2.charAt(j - 1)) { 
    					dp[i][j] = dp[i - 1][j - 1] + 1;
    				} else { 
    					dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
    				}
    			}
    		}
    
    		return dp[text1.length()][text2.length()];
    	}
    
    	// 方法二：方法一的空间优化
    	public int longestCommonSubsequence2(String s1, String s2) { 
    		int m = s1.length(), n = s2.length();
    		if (m < n)// 本判断语句目的可以减少空间复杂度。可以移除本判断语句。
    			return longestCommonSubsequence2(s2, s1);
    		int[][] dp = new int[2][n + 1];
    		for (int i = 1, k = 1; i <= m; ++i, k ^= 1)
    			for (int j = 1; j <= n; ++j)
    				if (s1.charAt(i - 1) == s2.charAt(j - 1))
    					dp[k][j] = 1 + dp[k ^ 1][j - 1];
    				else
    					dp[k][j] = Math.max(dp[k ^ 1][j], dp[k][j - 1]);
    		return dp[m & 1][n];
    	}
    
    	// 方法三：方法一的空间优化Plus
    	public int longestCommonSubsequence3(String text1, String text2) { 
    		int m = text1.length(), n = text2.length();
    		if (m < n) { // 本判断语句目的可以减少空间复杂度。可以移除本判断语句。
    			return longestCommonSubsequence3(text2, text1);
    		}
    
    		int[] dp = new int[n + 1];
    		for (int i = 1; i <= text1.length(); ++i) { 
    			for (int j = 1, left = 0, leftUp = 0, newOne = 0; j <= text2.length(); ++j) { 
    
    				if (text1.charAt(i - 1) == text2.charAt(j - 1)) { 
    					newOne = leftUp + 1;
    				} else { 
    					newOne = Math.max(left, dp[j]);
    				}
    				leftUp = dp[j];
    				left = dp[j] = newOne;// 为同行下一元素以及下行元素做准备
    
    			}
    		}
    		return dp[n];
    	}
    }

## Test ##

import static org.junit.Assert.*;
    import org.junit.Test;
    
    public class LongestCommonSubsequenceTest { 
    
    	@Test
    	public void test() { 
    		LongestCommonSubsequence obj = new LongestCommonSubsequence();
    
    		assertEquals(3, obj.longestCommonSubsequence("abcde", "ace"));
    		assertEquals(3, obj.longestCommonSubsequence("ace", "abcde"));
    		assertEquals(3, obj.longestCommonSubsequence("abc", "abc"));
    		assertEquals(0, obj.longestCommonSubsequence("abc", "def"));
    	}
    
    	@Test
    	public void test2() { 
    		LongestCommonSubsequence obj = new LongestCommonSubsequence();
    
    		assertEquals(3, obj.longestCommonSubsequence2("abcde", "ace"));
    		assertEquals(3, obj.longestCommonSubsequence2("ace", "abcde"));
    		assertEquals(3, obj.longestCommonSubsequence2("abc", "abc"));
    		assertEquals(0, obj.longestCommonSubsequence2("abc", "def"));
    	}
    
    	@Test
    	public void test3() { 
    		LongestCommonSubsequence obj = new LongestCommonSubsequence();
    
    		assertEquals(3, obj.longestCommonSubsequence3("abcde", "ace"));
    		assertEquals(3, obj.longestCommonSubsequence3("ace", "abcde"));
    		assertEquals(3, obj.longestCommonSubsequence3("abc", "abc"));
    		assertEquals(0, obj.longestCommonSubsequence3("abc", "def"));
    	}
    
    	@Test
    	public void testOther() { 
    		assertEquals(1 & 1, 1 % 2);
    		assertEquals(2 & 1, 2 % 2);
    		assertEquals(3 & 1, 3 % 2);
    		assertEquals(4 & 1, 4 % 2);
    	}
    }

[https_leetcode.com_problems_longest-common-subsequence]: https://leetcode.com/problems/longest-common-subsequence/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3UwMTE4NjMwMjQ_size_16_color_FFFFFF_t_70_pic_center]: /images/20221022/f696824dccfc42ebb7036e2ae831fc60.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3UwMTE4NjMwMjQ_size_16_color_FFFFFF_t_70_pic_center 1]: /images/20221022/7814040afdf74f47a255f3226929b487.png
[Link 1]: https://mp.weixin.qq.com/s/Qq0q4HaE4TyasCTj2WGFOg
[Java_Python 3_ Two DP codes of O_mn_ _ O_min_m_ n_ spaces w_ picture and analysis]: https://leetcode.com/problems/longest-common-subsequence/discuss/351689/JavaPython-3-Two-DP-codes-of-O%28mn%29-and-O%28min%28m-n%29%29-spaces-w-picture-and-analysis