1352. Product of the Last K Numbers

 

You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot.

The lock initially starts at '0000', a string representing the state of the 4 wheels.

You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.

Given a target representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.

Example 1:

Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
Output: 6
Explanation:
A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
because the wheels of the lock become stuck after the display becomes the dead end "0102".

 

Example 2:

Input: deadends = ["8888"], target = "0009"
Output: 1
Explanation:
We can turn the last wheel in reverse to move from "0000" -> "0009".

 

Example 3:

Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
Output: -1
Explanation:
We can't reach the target without getting stuck.

 

Example 4:

Input: deadends = ["0000"], target = "8888"
Output: -1

 

Note:

1. The length of deadends will be in the range [1, 500].
2. target will not be in the list deadends.
3. Every string in deadends and the string target will be a string of 4 digits from the 10,000 possibilities '0000' to '9999'.

 

这道题说有一种可滑动的四位数的锁，貌似行李箱上比较常见这种锁。给了我们一个目标值，还有一些死锁的情况，就是说如果到达这些死锁的位置，就不能再动了，相当于迷宫中的障碍物。然后问我们最少多少步可以从初始的0000位置滑动到给定的target位置。如果各位足够老辣的话，应该能发现其实本质就是个迷宫遍历的问题，只不过相邻位置不再是上下左右四个位置，而是四位数字每个都加一减一，总共有八个相邻的位置。遍历迷宫问题中求最短路径要用BFS来做，那么这道题也就是用BFS来解啦，和经典BFS遍历迷宫解法唯一不同的就是找下一个位置的地方，这里我们要遍历四位数字的每一位，然后分别加1减1，我们用j从-1遍历到1，遇到0跳过，也就是实现了加1减1的过程。然后我们要计算要更新位上的数字，为了处理9加1变0，和0减1变9的情况，我们统一给该位数字加上个10，然后再加或减1，最后再对10取余即可，注意字符和整型数之间通过加或减’0’来转换。我们用结果res来记录BFS遍历的层数，如果此时新生成的字符串等于target了，直接返回结果res，否则我们看如果该字符串不在死锁集合里，且之前没有遍历过，那么加入队列queue中，之后将该字符串加入visited集合中即可。注意这里在while循环中，由于要一层一层的往外扩展，一般的做法是会用一个变量len来记录当前的q.size()，博主为了简洁，使用了一个trick，就是从q.size()往0遍历，千万不能反回来，因为在计算的过程中q的大小会变化，如果让k < q.size() 为终止条件，绝b会出错，而我们初始化为q.size()就没事，参见代码如下：

 

解法一：

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        unordered_set<string> deadlock(deadends.begin(), deadends.end());
        if (deadlock.count("0000")) return -1;
        int res = 0;
        unordered_set<string> visited{{"0000"}};
        queue<string> q{{"0000"}};
        while (!q.empty()) {
            ++res;
            for (int k = q.size(); k > 0; --k) {
                auto t = q.front(); q.pop();
                for (int i = 0; i < t.size(); ++i) {
                    for (int j = -1; j <= 1; ++j) {
                        if (j == 0) continue;
                        string str = t;
                        str[i] = ((t[i] - '0') + 10 + j) % 10 + '0';
                        if (str == target) return res;
                        if (!visited.count(str) && !deadlock.count(str)) q.push(str);        
                        visited.insert(str);
                    }
                }
            }
        }
        return -1;
    }
};

 

下面这种方法也是用的BFS遍历，不同之处在于生成新字符串的方法，这里我们采用拼接法来生成新字符串，而不是像上面那样使用置换字符串的方法。我们对于加一和减一分别进行拼接，注意处理9加1变0，和0减1变9的情况。然后剩下的部分就和经典的BFS遍历写法没有什么太大的区别了，参见代码如下：

 

解法二：

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        unordered_set<string> deadlock(deadends.begin(), deadends.end());
        if (deadlock.count("0000")) return -1;
        int res = 0;
        unordered_set<string> visited{{"0000"}};
        queue<string> q{{"0000"}};
        while (!q.empty()) {
            ++res;
            for (int k = q.size(); k > 0; --k) {
                auto t = q.front(); q.pop();
                for (int i = 0; i < t.size(); ++i) {
                    char c = t[i];
                    string str1 = t.substr(0, i) + to_string(c == '9' ? 0 : c - '0' + 1) + t.substr(i + 1);
                    string str2 = t.substr(0, i) + to_string(c == '0' ? 9 : c - '0' - 1) + t.substr(i + 1);
                    if (str1 == target || str2 == target) return res;
                    if (!visited.count(str1) && !deadlock.count(str1)) q.push(str1);
                    if (!visited.count(str2) && !deadlock.count(str2)) q.push(str2);
                    visited.insert(str1);
                    visited.insert(str2);
                }
            }
        }
        return -1;
    }
};

 

参考资料：

https://leetcode.com/problems/open-the-lock

https://leetcode.com/problems/open-the-lock/discuss/110230/BFS-solution-C++

https://leetcode.com/problems/open-the-lock/discuss/110237/Regular-java-BFS-solution-and-2-end-BFS-solution-with-improvement

 

LeetCode All in One 题目讲解汇总(持续更新中…)

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