Question

583. 两个字符串的删除操作

English Version

题目描述

给定两个单词 word1 和

 word2 ，返回使得 word1 和  word2_ _相同所需的最小步数。

每步 可以删除任意一个字符串中的一个字符。

 

示例 1：

输入: word1 = "sea", word2 = "eat"
输出: 2
解释: 第一步将 "sea" 变为 "ea" ，第二步将 "eat "变为 "ea"

示例  2:

输入：word1 = "leetcode", word2 = "etco"
输出：4

 

提示：

• 1 <= word1.length, word2.length <= 500
• word1 和 word2 只包含小写英文字母

解法

方法一：动态规划

类似1143. 最长公共子序列。

定义 dp[i][j] 表示使得 word1[0:i-1] 和 word1[0:j-1] 两个字符串相同所需执行的删除操作次数。

时间复杂度：$O(mn)$。

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

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

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

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

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

impl Solution {
  pub fn min_distance(word1: String, word2: String) -> i32 {
    let (m, n) = (word1.len(), word2.len());
    let (word1, word2) = (word1.as_bytes(), word2.as_bytes());
    let mut dp = vec![vec![0; n + 1]; m + 1];
    for i in 1..=m {
      for j in 1..=n {
        dp[i][j] = if word1[i - 1] == word2[j - 1] {
          dp[i - 1][j - 1] + 1
        } else {
          dp[i - 1][j].max(dp[i][j - 1])
        };
      }
    }
    let max = dp[m][n];
    (m - max + (n - max)) as i32
  }
}