LeetCode解题分享：752. Open the Lock

Problem

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’.

解题思路

   典型的BFS题。对于这一类问题，往往需要较多的查找操作，而使用列表或者Vector之类的数据结构容易超时，因此一个小技巧就是将其转化为set，这样查找的速度会快很多。以下是一份可以通过的代码，够直观，但效率不是很高。

   代码如下：

class Solution {
public:
	int openLock(const vector<string>& deadends, string target) {
		if (find(deadends.begin(), deadends.end(), "0000") != deadends.end())
		{
			return -1;
		}
		set<string> trap(deadends.begin(), deadends.end());

		set<string> visited;

		queue<pair<string, int>> q;

		q.push(make_pair("0000", 0));

		while (!q.empty())
		{
			pair<string, int> node = q.front();
			q.pop();

			string s = node.first;
			int length = node.second;

			if (s == target)
			{
				return length;
			}
			for (int i = 0; i < 4; ++i)
			{
				string temp = s;
				temp[i] = (temp[i] - '0' + 1) % (10) + '0';
				if (trap.find(temp) == trap.end() && 
					visited.find(temp) == visited.end())
				{
					visited.insert(temp);
					q.push(make_pair(temp, length + 1));
				}
				temp = s;
				temp[i] = (temp[i] - '0' - 1 + 10) % (10) + '0';
				if (trap.find(temp) == trap.end() &&
					visited.find(temp) == visited.end())
				{
					visited.insert(temp);
					q.push(make_pair(temp, length + 1));
				}
			}
		}
		return -1;
	}
};

标签：752,0000,target,temp,Lock,deadends,end,Open,string
来源： https://blog.csdn.net/DaVinciL/article/details/99692867