Question

• tags: [ DP_Two_Sequence ]

Source

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

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] 和 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

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++

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

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⋅len2)O(len1 \cdot len2)O(len1⋅len2). 使用二维矩阵，空间复杂度为 O(len1⋅len2)O(len1 \cdot len2)O(len1⋅len2).