Question

72. 编辑距离

English Version

题目描述

给你两个单词 word1 和 word2， _请返回将 word1 转换成 word2 所使用的最少操作数_  。

你可以对一个单词进行如下三种操作：

• 插入一个字符
• 删除一个字符
• 替换一个字符

 

示例 1：

输入：word1 = "horse", word2 = "ros"
输出：3
解释：
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')

示例 2：

输入：word1 = "intention", word2 = "execution"
输出：5
解释：
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')

 

提示：

• 0 <= word1.length, word2.length <= 500
• word1 和 word2 由小写英文字母组成

解法

方法一：动态规划

我们定义 $f[i][j]$ 表示将 $word1$ 的前 $i$ 个字符转换成 $word2$ 的前 $j$ 个字符所使用的最少操作数。初始时 $f[i][0] = i$, $f[0][j] = j$。其中 $i \in [1, m], j \in [0, n]$。

考虑 $f[i][j]$：

• 如果 $word1[i - 1] = word2[j - 1]$，那么我们只需要考虑将 $word1$ 的前 $i - 1$ 个字符转换成 $word2$ 的前 $j - 1$ 个字符所使用的最少操作数，因此 $f[i][j] = f[i - 1][j - 1]$；
• 否则，我们可以考虑插入、删除、替换操作，那么 $f[i][j] = \min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1$。

综上，我们可以得到状态转移方程：

$$ f[i][j] = \begin{cases} i, & \text{if } j = 0 \ j, & \text{if } i = 0 \ f[i - 1][j - 1], & \text{if } word1[i - 1] = word2[j - 1] \ \min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1, & \text{otherwise} \end{cases} $$

最后，我们返回 $f[m][n]$ 即可。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是 $word1$ 和 $word2$ 的长度。

class Solution:
  def minDistance(self, word1: str, word2: str) -> int:
    m, n = len(word1), len(word2)
    f = [[0] * (n + 1) for _ in range(m + 1)]
    for j in range(1, n + 1):
      f[0][j] = j
    for i, a in enumerate(word1, 1):
      f[i][0] = i
      for j, b in enumerate(word2, 1):
        if a == b:
          f[i][j] = f[i - 1][j - 1]
        else:
          f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
    return f[m][n]

class Solution {
  public int minDistance(String word1, String word2) {
    int m = word1.length(), n = word2.length();
    int[][] f = new int[m + 1][n + 1];
    for (int j = 1; j <= n; ++j) {
      f[0][j] = j;
    }
    for (int i = 1; i <= m; ++i) {
      f[i][0] = i;
      for (int j = 1; j <= n; ++j) {
        if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
          f[i][j] = f[i - 1][j - 1];
        } else {
          f[i][j] = Math.min(f[i - 1][j], Math.min(f[i][j - 1], f[i - 1][j - 1])) + 1;
        }
      }
    }
    return f[m][n];
  }
}

class Solution {
public:
  int minDistance(string word1, string word2) {
    int m = word1.size(), n = word2.size();
    int f[m + 1][n + 1];
    for (int j = 0; j <= n; ++j) {
      f[0][j] = j;
    }
    for (int i = 1; i <= m; ++i) {
      f[i][0] = i;
      for (int j = 1; j <= n; ++j) {
        if (word1[i - 1] == word2[j - 1]) {
          f[i][j] = f[i - 1][j - 1];
        } else {
          f[i][j] = min({f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]}) + 1;
        }
      }
    }
    return f[m][n];
  }
};

func minDistance(word1 string, word2 string) int {
  m, n := len(word1), len(word2)
  f := make([][]int, m+1)
  for i := range f {
    f[i] = make([]int, n+1)
  }
  for j := 1; j <= n; j++ {
    f[0][j] = j
  }
  for i := 1; i <= m; i++ {
    f[i][0] = i
    for j := 1; j <= n; j++ {
      if word1[i-1] == word2[j-1] {
        f[i][j] = f[i-1][j-1]
      } else {
        f[i][j] = min(f[i-1][j], min(f[i][j-1], f[i-1][j-1])) + 1
      }
    }
  }
  return f[m][n]
}

function minDistance(word1: string, word2: string): number {
  const m = word1.length;
  const n = word2.length;
  const f: number[][] = Array(m + 1)
    .fill(0)
    .map(() => Array(n + 1).fill(0));
  for (let j = 1; j <= n; ++j) {
    f[0][j] = j;
  }
  for (let i = 1; i <= m; ++i) {
    f[i][0] = i;
    for (let j = 1; j <= n; ++j) {
      if (word1[i - 1] === word2[j - 1]) {
        f[i][j] = f[i - 1][j - 1];
      } else {
        f[i][j] = Math.min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1;
      }
    }
  }
  return f[m][n];
}

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function (word1, word2) {
  const m = word1.length;
  const n = word2.length;
  const f = Array(m + 1)
    .fill(0)
    .map(() => Array(n + 1).fill(0));
  for (let j = 1; j <= n; ++j) {
    f[0][j] = j;
  }
  for (let i = 1; i <= m; ++i) {
    f[i][0] = i;
    for (let j = 1; j <= n; ++j) {
      if (word1[i - 1] === word2[j - 1]) {
        f[i][j] = f[i - 1][j - 1];
      } else {
        f[i][j] = Math.min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1;
      }
    }
  }
  return f[m][n];
};