【leetcode】Edit Distance

最新推荐文章于 2019-04-16 22:44:18 发布

iCoder_Me

最新推荐文章于 2019-04-16 22:44:18 发布

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分类专栏：算法文章标签： c leetcode 算法动态规划

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本文介绍了一种计算两个字符串之间的编辑距离的算法，并提供了一个详细的C语言实现案例。该算法允许三种基本操作：插入字符、删除字符及替换字符，并通过递归结合动态规划的方式实现了高效的编辑距离计算。

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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:

a) Insert a character
b) Delete a character

c) Replace a character

Accept: 12ms

int dp[1000][1000];

int findDist(char *word1, size_t len1, char *word2, size_t len2) {
  if (dp[len1][len2] != 0) {
    return dp[len1][len2];
  }
  if (*word1 == '\0') {
    dp[0][len2] = len2;
    return dp[0][len2];
  } else if (*word2 == '\0') {
    dp[len1][0] = len1;
    return dp[len1][0];
  }

  if (*word1 == *word2) {
    dp[len1][len2] = findDist(word1+1, len1-1, word2+1, len2-1);
    return dp[len1][len2];
  }
  // REPLACE
  int d1 = 1 + findDist(word1+1, len1-1, word2+1, len2-1);
  // DELETE
  int d2 = 1 + findDist(word1+1, len1-1, word2, len2);
  // INSERT
  int d3 = 1 + findDist(word1, len1, word2+1, len2-1);

  int m = d1;
  if (d1 < d2) {
    if (d1 < d3) {
      m = d1;
    } else {
      m = d3;
    }
  } else {
    if (d2 < d3) {
      m = d2;
    } else {
      m = d3;
    }
  }
  dp[len1][len2] = m;
  return dp[len1][len2];
}

int minDistance(char *word1, char *word2) {
  size_t len1 = strlen(word1);
  size_t len2 = strlen(word2);
  for (size_t i = 0; i <= len1; ++i) {
    for (size_t j = 0; j <= len2; ++j) {
      dp[i][j] = 0;
    }
  }
  return findDist(word1, len1, word2, len2);
}