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编辑距离 #311

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Sunny-117 opened this issue Nov 8, 2022 · 1 comment
Open

编辑距离 #311

Sunny-117 opened this issue Nov 8, 2022 · 1 comment

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@Sunny-117
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@floatGray
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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')

var minDistance = function(word1, word2) {
  const dp = Array.from({ length: word2.length + 1 }, (item, index) => index);
  // 保存的是左上角的值
  let before;
  for (let i = 1; i <= word1.length; i++) {
    before = dp[ 0 ];
    dp[ 0 ] = i;
    for (let j = 1; j < dp.length; j++) {
      // 先将左上角值缓存,后续在进行更新
      const temp = dp[ j ];
      const flag = word1[i - 1] === word2[j - 1] ? 1 : 0;
      dp[j] = 1 + Math.min(dp[j - 1], dp[j], before - flag);
      before = temp;
    }
  }
  return dp[ dp.length - 1 ]
};

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