Edit Distance

Question

  • leetcode: Edit Distance | LeetCode OJ
  • lintcode: (119) Edit Distance
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    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:

    Insert a character
    Delete a character
    Replace a character
    Example
    Given word1 = "mart" and word2 = "karma", return 3.

题解1 - 双序列动态规划

两个字符串比较,求最值,直接看似乎并不能直接找出解决方案,这时往往需要使用动态规划的思想寻找递推关系。使用双序列动态规划的通用做法,不妨定义f[i][j]为字符串1的前i个字符和字符串2的前j个字符的编辑距离,那么接下来寻找其递推关系。增删操作互为逆操作,即增或者删产生的步数都是一样的。故初始化时容易知道f[0][j] = j, f[i][0] = i, 接下来探讨f[i][j]f[i - 1][j - 1]的关系,和 LCS 问题类似,我们分两种情况讨论,即word1[i] == word2[j] 与否,第一种相等的情况有:

  1. i == j, 且有word1[i] == word2[j], 则由f[i - 1][j - 1] -> f[i][j] 不增加任何操作,有f[i][j] = f[i - 1][j - 1].
  2. i != j, 由于字符数不等,肯定需要增/删一个字符,但是增删 word1 还是 word2 是不知道的,故可取其中编辑距离的较小值,即f[i][j] = 1 + min{f[i - 1][j], f[i][j - 1]}.

第二种不等的情况有:

  1. i == j, 有f[i][j] = 1 + f[i - 1][j - 1].
  2. i != j, 由于字符数不等,肯定需要增/删一个字符,但是增删 word1 还是 word2 是不知道的,故可取其中编辑距离的较小值,即f[i][j] = 1 + min{f[i - 1][j], f[i][j - 1]}.

最后返回f[len(word1)][len(word2)]

Python

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class Solution: 
# @param word1 & word2: Two string.
# @return: The minimum number of steps.
def minDistance(self, word1, word2):
len1, len2 = 0, 0
if word1:
len1 = len(word1)
if word2:
len2 = len(word2)
if not word1 or not word2:
return max(len1, len2)

f = [[i + j for i in xrange(1 + len2)] for j in xrange(1 + len1)]

for i in xrange(1, 1 + len1):
for j in xrange(1, 1 + len2):
if word1[i - 1] == word2[j - 1]:
f[i][j] = min(f[i - 1][j - 1], 1 + f[i - 1][j], 1 + f[i][j - 1])
else:
f[i][j] = 1 + min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1])
return f[len1][len2]

C++

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class Solution {
public:
/**
* @param word1 & word2: Two string.
* @return: The minimum number of steps.
*/
int fistance(string word1, string word2) {
if (word1.empty() || word2.empty()) {
return max(word1.size(), word2.size());
}

int len1 = word1.size();
int len2 = word2.size();
vector<vector<int> > f = \
vector<vector<int> >(1 + len1, vector<int>(1 + len2, 0));
for (int i = 0; i <= len1; ++i) {
f[i][0] = i;
}
for (int i = 0; i <= len2; ++i) {
f[0][i] = i;
}

for (int i = 1; i <= len1; ++i) {
for (int j = 1; j <= len2; ++j) {
if (word1[i - 1] == word2[j - 1]) {
f[i][j] = min(f[i - 1][j - 1], 1 + f[i - 1][j]);
f[i][j] = min(f[i][j], 1 + f[i][j - 1]);
} else {
f[i][j] = min(f[i - 1][j - 1], f[i - 1][j]);
f[i][j] = 1 + min(f[i][j], f[i][j - 1]);
}
}
}

return f[len1][len2];
}
};

Java

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public class Solution {
public int minDistance(String word1, String word2) {
int len1 = 0, len2 = 0;
if (word1 != null && word2 != null) {
len1 = word1.length();
len2 = word2.length();
}
if (word1 == null || word2 == null) {
return Math.max(len1, len2);
}

int[][] f = new int[1 + len1][1 + len2];
for (int i = 0; i <= len1; i++) {
f[i][0] = i;
}
for (int i = 0; i <= len2; i++) {
f[0][i] = i;
}

for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
f[i][j] = Math.min(f[i - 1][j - 1], 1 + f[i - 1][j]);
f[i][j] = Math.min(f[i][j], 1 + f[i][j - 1]);
} else {
f[i][j] = Math.min(f[i - 1][j - 1], f[i - 1][j]);
f[i][j] = 1 + Math.min(f[i][j], f[i][j - 1]);
}
}
}

return f[len1][len2];
}
}

源码解析

  1. 边界处理
  2. 初始化二维矩阵(Python 中初始化时 list 中 len2 在前,len1 在后)
  3. i, j 从1开始计数,比较 word1 和 word2 时注意下标
  4. 返回f[len1][len2]

复杂度分析

两重 for 循环,时间复杂度为 $$O(len1 \cdot len2)$$. 使用二维矩阵,空间复杂度为 $$O(len1 \cdot len2)$$.




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