编辑距离和最长公共子串

编辑距离和最长公共子串问题都是经典的DP问题，首先来看看编辑距离问题：

问题描述

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

解决思路

经典的动态规划题，建立一个二维的数组dp[][]记录两个字符串s1和s2子串的最短编辑距离，递推公式如下：

(1) 当s1.charAt(i) == s2.charAt(j)时，dp[i][j] = dp[i - 1][j - 1];

(2) 其他时，dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j]));

初始化条件为：

dp[i][0] = i, dp[0][j] = j;

代码

	int getMinEditLen(String s1, String s2) {
		if (s1 == null && s2 == null) {
			return 0;
		}

		if (s1.length() == 0) {
			return s2.length();
		}
		if (s2.length() == 0) {
			return s1.length();
		}

		int len1 = s1.length();
		int len2 = s2.length();

		int[][] dp = new int[len1 + 1][len2 + 1];
		// initialize
		for (int i = 0; i < dp.length; i++) {
			dp[i][0] = i;
		}
		for (int j = 0; j < dp[0].length; j++) {
			dp[0][j] = j;
		}

		for (int i = 1; i < dp.length; i++) {
			for (int j = 1; j < dp[0].length; j++) {
				if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
					dp[i][j] = dp[i - 1][j - 1];
				} else {
					dp[i][j] = Math.min(dp[i - 1][j - 1],
							Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
				}
			}
		}

		return dp[len1][len2];
	}

 容易写错的地方

if (word1.substring(0, i).equals(word2.substring(0, j))) {
        dp[i][j] = 0;
}

最长公共子串问题

问题描述

子字符串的定义和子序列的定义类似，但要求是连续分布在其他字符串中。比如输入两个字符串BDCABA和ABCBDAB的最长公共字符串有BD和AB，它们的长度都是2。

解决思路

(1) 递归；

(2) dp;

代码

	// rec
	int getLCS(String s1, String s2) {
		if (s1.length() == 0 || s2.length() == 0) {
			return 0;
		}

		int len1 = s1.length();
		int len2 = s2.length();

		if (s1.charAt(len1 - 1) == s2.charAt(len2 - 1)) {
			return getLCS(s1.substring(0, len1 - 1), s2.substring(0, len2 - 1)) + 1;
		}

		return Math.max(
				getLCS(s1.substring(0, len1), s2.substring(0, len2 - 1)),
				getLCS(s1.substring(0, len1 - 1), s2.substring(0, len2)));
	}

	// dp
	int getLCS(String s1, String s2) {
		if (s1.length() == 0 || s2.length() == 0) {
			return 0;
		}

		int len1 = s1.length();
		int len2 = s2.length();

		int[][] dp = new int[len1 + 1][len2 + 1];
		for (int i = 1; i < dp.length; i++) {
			for (int j = 1; j < dp[0].length; j++) {
				if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
					dp[i][j] = dp[i - 1][j - 1] + 1;
				} else {
					dp[i][j] = 0;
				}
			}
		}

		return dp[len1][len2];
	}

拓展：最公共子序列问题。

例如s1 = "abc", s2 = "asbvcd", s1和s2的最长公共子序列为"abc"，长度为3.

public class LongestCommonSequence {
	public int getLCSeqLen(String s1, String s2) {
		if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) {
			return 0;
		}

		int len1 = s1.length(), len2 = s2.length();
		int[][] dp = new int[len1 + 1][len2 + 1];

		for (int i = 1; i <= len1; i++) {
			for (int j = 1; j <= len2; j++) {
				if (s1.charAt(i - 1) == s2.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[len1][len2];
	}
	
	public String getLCSeq(String s1, String s2) {
		if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) {
			return "";
		}

		int len1 = s1.length(), len2 = s2.length();
		int[][] dp = new int[len1 + 1][len2 + 1];
		
		String lcs = "";
		
		for (int i = 1; i <= len1; i++) {
			for (int j = 1; j <= len2; j++) {
				if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
					dp[i][j] = dp[i - 1][j - 1] + 1;
					lcs += s1.charAt(i - 1);
				} else {
					dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
				}
			}
		}
		
		return lcs;
	}
}