[LeetCode] 0752. Open the Lock 打开转盘锁
题目
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.
你有一个带有四个圆形拨轮的转盘锁。每个拨轮都有10个数字： '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' 。每个拨轮可以自由旋转：例如把 '9' 变为 '0'，'0' 变为 '9' 。每次旋转都只能旋转一个拨轮的一位数字。
锁的初始数字为 '0000' ，一个代表四个拨轮的数字的字符串。
列表 deadends 包含了一组死亡数字，一旦拨轮的数字和列表里的任何一个元素相同，这个锁将会被永久锁定，无法再被旋转。
字符串 target 代表可以解锁的数字，你需要给出最小的旋转次数，如果无论如何不能解锁，返回 -1。
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:
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'.
解法一
思路：就是一道简单的BFS搜索题，要注意的点就是每次搜索时，要先遍历队列中的所有元素，全部处理完了再做第二轮搜索。还有就是在遍历队列并修改时，要先把size存下来，否则会得到错误的结果。
不足：不知道为什么运行速度和内存使用都这么大，一方面可能是计算移动时直接用的字符操作，效果没有用int来的快，一方面是visted用的unordered_set<string>，效率和内存肯定是没有bool visited[10000] + memset(visited, 0, sizeof(visited))来的好。
重新刷leetcode的第二天，先这样实现了练练手，之后再做性能改进吧。
class Solution {
public:
int openLock(vector<string>& deadends, string target) {
unordered_set<string> deadset(deadends.begin(), deadends.end());
if (deadset.count("0000")) {
return -1;
}
queue<string> q;
unordered_set<string> visited;
q.push("0000");
visited.insert("0000");
int result = 0;
// 开始BFS搜索
while (!q.empty()) {
result++;
auto size = q.size();
for (int i = 0; i < size; i++) {
string seq = q.front();
q.pop();
// 获取所有有效移动
auto moves = getValidMoves(seq);
for (auto& move : moves) {
if (move == target) {
return result;
}
// 如果该移动不是deadend且没访问过，则入队
if (!deadset.count(move) && !visited.count(move)) {
q.push(move);
visited.insert(move);
}
// 如果是deadend则不做处理，相当于绕过deadend
}
}
}
// 没有找到目标序列，返回-1
return -1;
}
vector<string> getValidMoves(const string& sequence) {
vector<string> moves;
for (int i = 0; i < 4; i++) {
string temp = sequence;
// +1
temp[i] = temp[i] == '9' ? '0' : temp[i] + 1;
moves.push_back(temp);
// -1
temp = sequence;
temp[i] = temp[i] == '0' ? '9' : temp[i] - 1;
moves.push_back(temp);
}
return moves;
}
};
运行结果：
Runtime: 324 ms, faster than 27.97% of C++ online submissions for Open the Lock.
Memory Usage: 106.9 MB, less than 17.31% of C++ online submissions for Open the Lock.
解法二
可以说的上非常神奇了，首先仔细观察题目，可以发现以下几点：
- 目标不可达的条件是，deadends里必须有只和target差一步的序列（例如deadends = "8888", target = "8889"），也就是说，可以直接从deadends中看出target是否可达；
- 直接计算所有可达结果（target的前一步）的所需步长，并计算最小步长即可，根本不需要用到BFS；
这个算法真是可遇不可求呀~
class Solution {
public:
int openLock(vector<string>& deadends, string target) {
unordered_set<string> deadset(deadends.begin(), deadends.end());
if (deadset.count("0000") || deadset.count(target)) {
return -1;
}
vector<string> movesToTarget;
auto moves = getValidMoves(target);
for (auto& move : moves) {
if (!deadset.count(move)) {
movesToTarget.push_back(move);
}
}
// 可以直接从deadends中看出target可不可达
if (movesToTarget.empty()) {
return -1;
}
// 最大步长是40步（每位转动10次）
int min_stride = 40;
// 计算到达每个可达结果的步长，取最小
for (auto& move : movesToTarget) {
int cur_stride = 0;
for (int i = 0; i < 4; ++i) {
int turns = move[i] - '0';
// 可以倒着转，所以转动次数不会大过5
if (turns > 5) {
turns = 10 - turns;
}
cur_stride += turns;
}
if (cur_stride < min_stride) {
min_stride = cur_stride;
}
}
// 最后加上到达target的那一步
return min_stride + 1;
}
vector<string> getValidMoves(const string& sequence) {
vector<string> moves;
for (int i = 0; i < 4; i++) {
string temp = sequence;
// +1
temp[i] = temp[i] == '9' ? '0' : temp[i] + 1;
moves.push_back(temp);
// -1
temp = sequence;
temp[i] = temp[i] == '0' ? '9' : temp[i] - 1;
moves.push_back(temp);
}
return moves;
}
};
运行结果：
Runtime: 8 ms, faster than 100.00% of C++ online submissions for Open the Lock.
Memory Usage: 10.3 MB, less than 98.08% of C++ online submissions for Open the Lock.
