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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
public 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
private static 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
private static 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] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[len1][len2];
}
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原文地址：http://www.cnblogs.com/harrygogo/p/4588587.html
