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[LeetCode] 72. Edit Distance #72

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 72. Edit Distance #72

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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grandyang commented May 30, 2019


请点击下方图片观看讲解视频
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Given two strings word1 and word2, return the minimum number of operations required to convertword1 to word2.

You have the following three operations permitted on a word:

  • Insert a character
  • Delete a character
  • Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

Constraints:

  • 0 <= word1.length, word2.length <= 500
  • word1 and word2 consist of lowercase English letters.

这道题让求从一个字符串转变到另一个字符串需要的变换步骤,共有三种变换方式,插入一个字符,删除一个字符,和替换一个字符。题目乍眼一看并不难,但是实际上却暗藏玄机,对于两个字符串的比较,一般都会考虑一下用 HashMap 统计字符出现的频率,但是在这道题却不可以这么做,因为字符串的顺序很重要。还有一种比较常见的错误,就是想当然的认为对于长度不同的两个字符串,长度的差值都是要用插入操作,然后再对应每位字符,不同的地方用修改操作,但是其实这样可能会多用操作,因为删除操作有时同时可以达到修改的效果。比如题目中的例子1,当把 horse 变为 rorse 之后,之后只要删除第二个r,跟最后一个e,就可以变为 ros。实际上只要三步就完成了,因为删除了某个字母后,原来左右不相连的字母现在就连一起了,有可能刚好组成了需要的字符串。所以在比较的时候,要尝试三种操作,因为谁也不知道当前的操作会对后面产生什么样的影响。

对于当前比较的两个字符 word1[i] 和 word2[j],若二者相同,一切好说,直接跳到下一个位置。若不相同,有三种处理方法,首先是直接在 word1 中插入一个 word2[j],那么 word2 中的 word2[j] 位置的字符就跳过了,接着比较 word1[i] 和 word2[j+1] 即可。第二个种方法是删除,即将 word1[i] 字符直接删掉,接着比较 word1[i+1] 和 word2[j] 即可。第三种则是将 word1[i] 修改为 word2[j],接着比较 word1[i+1] 和 word[j+1] 即可。分析到这里,就可以直接写出递归的代码,但是很可惜会 Time Limited Exceed,所以必须要优化时间复杂度,需要去掉大量的重复计算,这里使用记忆数组 memo 来保存计算过的状态,从而可以通过 OJ,注意这里的 insertCnt,deleteCnt,replaceCnt 仅仅是表示当前对应的位置分别采用了插入,删除,和替换操作,整体返回的最小距离,后面位置的还是会调用递归返回最小的,参见代码如下:

解法一:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vector<vector<int>> memo(m, vector<int>(n));
        return dfs(word1, 0, word2, 0, memo);
    }
    int dfs(string& word1, int i, string& word2, int j, vector<vector<int>>& memo) {
        int m = word1.size(), n = word2.size(), res = 0;
        if (i == m) return n - j;
        if (j == n) return m - i;
        if (memo[i][j] > 0) return memo[i][j];
        if (word1[i] == word2[j]) {
            return dfs(word1, i + 1, word2, j + 1, memo);
        } else {
            int insertCnt = dfs(word1, i, word2, j + 1, memo);
            int deleteCnt = dfs(word1, i + 1, word2, j, memo);
            int replaceCnt = dfs(word1, i + 1, word2, j + 1, memo);
            res = min(insertCnt, min(deleteCnt, replaceCnt)) + 1;
        }
        return memo[i][j] = res;
    }
};

根据以往的经验,对于字符串相关的题目且求极值的问题,十有八九都是用动态规划 Dynamic Programming 来解,这道题也不例外。其实解法一的递归加记忆数组的方法也可以看作是 DP 的递归写法。这里需要维护一个二维的数组 dp,其大小为 mxn,m和n分别为 word1 和 word2 的长度。dp[i][j] 表示从 word1 的前i个字符转换到 word2 的前j个字符所需要的步骤。先给这个二维数组 dp 的第一行第一列赋值,这个很简单,因为第一行和第一列对应的总有一个字符串是空串,于是转换步骤完全是另一个字符串的长度。跟以往的 DP 题目类似,难点还是在于找出状态转移方程,可以举个例子来看,比如 word1 是 "bbc",word2 是 "abcd",可以得到 dp 数组如下:

  Ø a b c d
Ø 0 1 2 3 4
b 1 1 1 2 3
b 2 2 1 2 3
c 3 3 2 1 2

通过观察可以发现,当 word1[i] == word2[j] 时,dp[i][j] = dp[i - 1][j - 1],其他情况时,dp[i][j] 是其左,左上,上的三个值中的最小值加1,那么可以得到状态转移方程为:

dp[i][j] = / dp[i - 1][j - 1] if word1[i - 1] == word2[j - 1]

\ min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1 else

解法二:

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 = 0; i <= m; ++i) dp[i][0] = i;
        for (int i = 0; i <= n; ++i) dp[0][i] = i;
        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] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[m][n];
    }
};

Github 同步地址:

#72

类似题目:

One Edit Distance

Delete Operation for Two Strings

Minimum ASCII Delete Sum for Two Strings

Uncrossed Lines

Minimum White Tiles After Covering With Carpets

参考资料:

https://leetcode.com/problems/edit-distance/

https://leetcode.com/problems/edit-distance/discuss/25846/C%2B%2B-O(n)-space-DP

LeetCode All in One 题目讲解汇总(持续更新中...)

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