算法练习（16） —— Open the Lock

算法练习（16） —— Open the Lock

习题

本题取自 leetcode 中的

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栏目中的第752题：

Open the Lock

题目如下：

Description

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.

Note

The length of

deadends

will be in the range

[1, 500]

.

target

will not be in the list

deadends

.

Every string in

deadends

and the string

target

will be a string of 4 digits from the 10,000 possibilities

'0000'

to

'9999'

.

Example1

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”.

Example2

Input: deadends = [“8888”], target = “0009”

Output: 1

Explanation:

We can turn the last wheel in reverse to move from “0000” -> “0009”.

Example3

Input: deadends = [“8887”,”8889”,”8878”,”8898”,”8788”,”8988”,”7888”,”9888”], target = “8888”

Output: -1

Explanation:

We can’t reach the target without getting stuck.

Example4

Input: deadends = [“0000”], target = “8888”

Output: -1

思考与代码

首先理解题意，锁的每个数字可以从0到9自由转动，其中0和9相连，从9往下转就得到了0。默认从

"0000"

开始，要求算转到目标数字的最少数目。其中还有一些死锁点，一旦转到那些数字，锁就坏了，所以必须要避开那些点。当然如果无法到达就返回-1.

首先可以确定，每一次转动数字的时候，有2种转法：往前转和往后转。而一共四个数字，所以就有2*4种转法。很自然的可以想到用广搜的思想，对每个从队列中弹出的点分别尝试这些转法，可行的话插入列队尾，步数+1。最终自然能找到最小的步数。

这里除了注意算法之外，还要注意一些超时的问题。首先是创建新的vector用于存储转法时，可以采用拷贝函数的方法。其次是容器选择方面，用vector的迭代器作find，速度会过慢，导致超时…这时候可以采用c++11中的unordered_set，它是基于哈希表的，使用find函数的时候速度会提升超级多。如果是java语言的话，就可以采用hashset。不得不说，论做题的3大困扰，RE，WA，TLE，还是TLE最惹人烦吧

代码如下：

#include <string>
#include <vector>
#include <queue>
using namespace std;

class Solution {
private:
// find if the input string does in the vector
bool findString(vector<string>& strings, string s) {
vector<string>::iterator res = find(strings.begin(), strings.end(), s);
if (res == strings.end())
return false;
return true;
}

// get neighbor strings of the input string
vector<string> neighbors(string s) {
vector<string> res;
for (int i = 0; i < 4; i++) {
string temp = s;
temp[i] = (s[i] - '0' + 1) % 10 + '0';
res.push_back(temp);
temp[i] = (s[i] - '0' + 9) % 10 + '0';
res.push_back(temp);
}
return res;
}

public:
int openLock(vector<string>& deadends, string target) {
// initialize
vector<string> visited;
queue<string> q;

int steps = 0;
string start = "0000";
visited.push_back(start);
q.push(start);

while (!q.empty()) {
int len = q.size();
for (int i = 0; i < len; i++) {
string top = q.front();
q.pop();
vector<string> neigh(neighbors(top));
for (auto s : neigh) {
if (s == target)
return ++steps;
if (findString(visited, s))
continue;
if (!findString(deadends, s)) {
q.push(s);
visited.push_back(s);
}
}
}
steps++;
}

return -1;
}
};