Leetcode 752 Open the Lock

题目原文

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'.

题意分析

转自 ： here

从初始状态“0000”开始转动，每次只能将其中一位转动一格，可正转可反转，同时deadends中有多个状态，如果转到该状态则不能继续，也即必不可转到目标结果。求问转到目标结果的最小次数。

解法分析

本题需要找到一个最小次数，通常需要运用BFS，在BFS中加入剪枝，并且对于到过的状态建立unordered_set，通过上述方法减少遍历次数。C++代码如下：

class Solution {
public:
   /* static bool isEnd(string a,vector<string>& deadends){
        for(auto k:deadends){
            if(a==k)
                return 1;
        }
        return 0;
    }*/
    int openLock(vector<string>& deadends, string target) {
        queue<string> myQue;
        unordered_set<string> dde(deadends.begin(),deadends.end());
        if(dde.count("0000"))
            return -1;
        myQue.push("0000");
        int queSize=1;
        int layer=0;
        int i,j;
        string temp,temp1;
        unordered_set<string> mySet;
        mySet.insert("0000");
        while(!myQue.empty()){
            layer++;
            queSize=myQue.size();
            for(i=0;i<queSize;i++){
                temp=myQue.front();
                for(j=0;j<4;j++){
                    temp1=temp;//add
                    if(temp1[j]=='9')
                        temp1[j]='0';
                    else
                        temp1[j]+=1;
                    if(temp1==target)
                        return layer;
                    if(!dde.count(temp1)&&!mySet.count(temp1)){            
                            myQue.push(temp1);
                            mySet.insert(temp1);                            
                    }
                    temp1=temp;//sub
                    if(temp1[j]=='0')
                        temp1[j]='9';
                    else
                        temp1[j]-=1;
                    if(temp1==target)
                        return layer;
                    if(!dde.count(temp1)&&!mySet.count(temp1)){
                            myQue.push(temp1);
                            mySet.insert(temp1);                   
                    }
                }
                myQue.pop(); 
            }
        }
        return -1;  
    }
};

上述代码不超时的关键除了剪枝和记录去过的状态以外，还在于检测一个状态是否deadend的方法，如果采用每次都遍历deadend一遍的方法则会超时，如果将deadend做成一个unordered_set，则可以是查询复杂度从O(n)降到O(1)。