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  • [LeetCode] 936. Stamping The Sequence 戳印序列

  • You want to form a `target` string of lowercase letters.

    At the beginning, your sequence is target.length '?' marks.  You also have a stamp of lowercase letters.

    On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp.  You can make up to 10 * target.length turns.

    For example, if the initial sequence is "?????", and your stamp is "abc",  then you may make "abc??", "?abc?", "??abc" in the first turn.  (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)

    If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn.  If the sequence is not possible to stamp, return an empty array.

    For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".

    Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves.  Any answers specifying more than this number of moves will not be accepted.

    Example 1:

    Input: stamp = "abc", target = "ababc"
    Output: [0,2]
    ([1,0,2] would also be accepted as an answer, as well as some other answers.)
    

    Example 2:

    Input: stamp = "abca", target = "aabcaca"
    Output: [3,0,1]
    

    Note:

    1. 1 <= stamp.length <= target.length <= 1000
    2. stamp and target only contain lowercase letters.

    这道题给了一个目标字符串 target,还有一个印戳字符串 stamp,现在有一个长度跟 target 一样的一排问号,每次可以在某个位置盖上印戳,新的印戳将会覆盖之前的字符,不论是问号还是其他字符,现在让找出所有盖印戳的位置,使得刚好可以盖出给定的字符串 target。这道题乍一看感觉还挺难下手的,毕竟一排问号,我们怎么知道该从哪里开始盖,但是如果换一个方向,假如给的是 target 字符串,每次盖印章,将对应的位置变成星号,只要将 target 中所有的字符盖成星号,最终再把盖印章的顺序翻转一下,就是题目中所求了。这里参考的是 [votrubac 大神的帖子](https://leetcode.com/problems/stamping-the-sequence/discuss/189576/C%2B%2B-simple-greedy),比如 target="aabccbc",stamp="abc",那么首先肯定是在 target 中找整个的 abc,可以找到,从位置1出开始盖,target 变为 a***cbc,同时标记此时已经盖了3个字母,加入到 total 变量中。然后继续找 abc,没有的话,就需要改变印戳了,开始往里面加星号,首先加一个星号,加的位置有三个,分别是 ab*, a*c, *bc,发现这三种都无法匹配,于是开始加两个星号,就有 a**,**c,其中 a** 可以成功匹配,起始位置为0,total 加1,于是 target 变为 ****cbc,然后发现此时 **c 也可以成功匹配,起始位置为2,total 加1,target 变为 *****bc。现在并不需要给 stamp 中加三个星号,这样没有意义,要做的是再从开头找一遍,此时发现 *bc 可以匹配,起始位置为4,total 加2,到现在位置 target 完全变为星号,当无法进行盖印戳的时候,就退出循环,需要一个 isStamped 的变量来标记一下是否进行了戳印。循环退出后要将 res 数组翻转一下,同时还要看 total 是否等于 target 的长度,只有相等了,才说明每个字母都被戳印了,否则返回空集,参见代码如下:
    class Solution {
    public:
        vector<int> movesToStamp(string stamp, string target) {
            vector<int> res;
            int n = stamp.size(), total = 0;
            while (true) {
                bool isStamped = false;
                for (int size = n; size > 0; --size) {
                    for (int i = 0; i <= n - size; ++i) {
                        string t = string(i, '*') + stamp.substr(i, size) + string(n - size - i, '*');
                        auto pos = target.find(t);
                        while (pos != string::npos) {
                            res.push_back(pos);
                            isStamped = true;
                            total += size;
                            fill(begin(target) + pos, begin(target) + pos + n, '*');
                            pos = target.find(t);
                        }
                    }
                }
                if (!isStamped) break;
            }
            reverse(res.begin(), res.end());
            return total == target.size() ? res : vector<int>();
        }
    };
    

    Github 同步地址:

    https://github.com/grandyang/leetcode/issues/936


    参考资料:

    https://leetcode.com/problems/stamping-the-sequence/

    https://leetcode.com/problems/stamping-the-sequence/discuss/189576/C%2B%2B-simple-greedy

    https://leetcode.com/problems/stamping-the-sequence/discuss/189258/C%2B%2B-Reverse-Operation-30-ms-better-than-DFS

    https://leetcode.com/problems/stamping-the-sequence/discuss/450930/C%2B%2B-Reverse-the-stamping-process-%2B-Wildcard-matching


    [LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)
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  • 原文:https://www.cnblogs.com/grandyang/p/12501661.html
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