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[LeetCode] Delete Operation for Two Strings 两个字符串的删除操作

 

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

 

Note:

  1. The length of given words won't exceed 500.
  2. Characters in given words can only be lower-case letters.

 

这道题给了我们两个单词,问我们最少需要多少步可以让两个单词相等,每一步我们可以在任意一个单词中删掉一个字符。那么我们分析怎么能让步数最少呢,是不是知道两个单词最长的相同子序列的长度,并乘以2,被两个单词的长度之和减,就是最少步数了。其实这道题就转换成求Longest Common Subsequence最长相同子序列的问题,令博主意外的是,LeetCode中竟然没有这道题,这与包含万物的LeetCode的作风不符啊。不过没事,有这道题也行啊,对于这种玩字符串,并且是求极值的问题,十有八九都是用dp来解的,曾经有网友问博主,如何确定什么时候用greedy,什么时候用dp?其实博主也不不太清楚,感觉dp要更tricky一些,而且出现的概率大,所以博主一般会先考虑dp,如果实在想不出递推公式,那么就想想greedy能做不。如果有大神知道更好的区分方法,请一定留言告知博主啊,多谢!那么决定了用dp来做,就定义一个二维的dp数组,其中dp[i][j]表示word1的前i个字符和word2的前j个字符组成的两个单词的最长公共子序列的长度。下面来看递推式dp[i][j]怎么求,首先来考虑dp[i][j]和dp[i-1][j-1]之间的关系,我们可以发现,如果当前的两个字符相等,那么dp[i][j] = dp[i-1][j-1] + 1,这不难理解吧,因为最长相同子序列又多了一个相同的字符,所以长度加1。由于我们dp数组的大小定义的是(n1+1) x (n2+1),所以我们比较的是word1[i-1]和word2[j-1]。那么我们想如果这两个字符不相等呢,难道我们直接将dp[i-1][j-1]赋值给dp[i][j]吗,当然不是,我们还要错位相比嘛,比如就拿题目中的例子来说,"sea"和"eat",当我们比较第一个字符,发现's'和'e'不相等,下一步就要错位比较啊,比较sea中第一个's'和eat中的'a',sea中的'e'跟eat中的第一个'e'相比,这样我们的dp[i][j]就要取dp[i-1][j]跟dp[i][j-1]中的较大值了,最后我们求出了最大共同子序列的长度,就能直接算出最小步数了,参见代码如下:

 

解法一:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++j) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return n1 + n2 - 2 * dp[n1][n2];
    }
};

 

下面这种方法也是用的dp,但是和上面的dp思路不太一样,这种算法是跟之前那道Edit Distance相同的思路。那道题问我们一个单词通过多少步修改可以得到另一个单词,其实word2删除一个字符,和跟在word1对应的地方加上那个要删除的字符,达到的效果是一样的,并不影响最终的步骤数,所以这道题完全可以按照那道题的解法来做,一点都不需要变动,定义一个二维的dp数组,其中dp[i][j]表示word1的前i个字符和word2的前j个字符组成的两个单词,能使其变相同的最小的步数,讲解可以参看那篇帖子,参见代码入下:

 

解法二:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
        for (int i = 0; i <= n1; ++i) dp[i][0] = i;
        for (int j = 0; j <= n2; ++j) dp[0][j] = j;
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++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[n1][n2];
    }
};

 

下面这种方法是解法二的递归写法,用的优化的dfs的方法,用memo数组来保存中间计算结果,以避免大量的重复计算,参见代码如下:

 

解法三:

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        vector<vector<int>> memo(n1 + 1, vector<int>(n2 + 1, 0));
        return helper(word1, word2, 0, 0, memo);
    }
    int helper(string word1, string word2, int p1, int p2, vector<vector<int>>& memo) {
        if (memo[p1][p2] != 0) return memo[p1][p2];
        int n1 = word1.size(), n2 = word2.size();
        if (p1 == n1 || p2 == n2) return n1 - p1 + n2 - p2;
        if (word1[p1] == word2[p2]) {
            memo[p1][p2] = helper(word1, word2, p1 + 1, p2 + 1, memo);
        } else {
            memo[p1][p2] = 1 + min(helper(word1, word2, p1 + 1, p2, memo), helper(word1, word2, p1, p2 + 1, memo));
        }
        return memo[p1][p2];
    }
};

 

类似题目:

Edit Distance

Minimum ASCII Delete Sum for Two Strings

 

参考资料:

https://discuss.leetcode.com/topic/89285/java-dp-solution-longest-common-subsequence

https://discuss.leetcode.com/topic/89308/dynamic-programming-and-memoization-solution

https://discuss.leetcode.com/topic/89433/two-pointers-recursive-java-solution-with-memoization

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2017-07-10 06:53  Grandyang  阅读(...)  评论(...编辑  收藏
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