leetcode第72题编辑距离

题目： 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 题目链接

我的JavaScript解法

• 解法一: 备忘录动态规划

var minDistance = function(word1, word2) {
  const m = word1.length;
  const n = word2.length;
  let memo = new Array();
  for(let i = 0; i < m; i++)
    memo[i] = new Array(n).fill(-1);
  return dp(word1, m-1, word2, n-1, memo);
};

const dp = (word1, i, word2, j, memo) => {
  if (i == -1) return j+1;
  if (j == -1) return i+1;
  if (memo[i][j] != -1)
    return memo[i][j];
  
  if (word1[i] == word2[j]) {
    memo[i][j] = dp(word1, i - 1, word2, j - 1, memo);
  } else {
    memo[i][j] = Math.min(
      dp(word1, i, word2, j - 1, memo) + 1,
      dp(word1, i - 1, word2, j, memo) + 1,
      dp(word1, i - 1, word2, j - 1, memo) + 1
    );
  }
  return memo[i][j];
}

• 解法二： DP table动态规划

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function(word1, word2) {
  const m = word1.length;
  const n = word2.length;
  let dp = new Array(m+1);
  for(let i = 0; i <= m; i++)
    dp[i] = new Array(n+1).fill(0);
  for(let i = 1; i<=m; i++)
    dp[i][0] = i;
  for(let j = 1; j<=n; j++)
    dp[0][j] = j;

  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];
      else
        dp[i][j] = Math.min(dp[i-1][j] + 1, dp[i][j-1]+1, dp[i-1][j-1]+1);
    }
  return dp[m][n];
}