LWC 64: 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:
- 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’.
思路：
实际上每一位有10种变化情况，总共有4位，所以有410种状态，初始从”0000”开始衍变，到target的最短路径可以用BFS来表达。当然，当衍变到deadends中的某些状态时，就不能再衍变了。典型的BFS+状态记录。
Java版本：
public int openLock(String[] deadends, String target) {
Set<String> vis = new HashSet<>();
for (int i = 0; i < deadends.length; ++i) {
vis.add(deadends[i]);
}
Queue<String> queue = new ArrayDeque<>();
queue.offer("0000");
if (vis.contains("0000")) return -1;
vis.add("0000");
int[] downUp = {1, -1};
int turn = 0;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; ++i) {
String now = queue.poll();
if (now.equals(target)) {
return turn;
}
for (int j = 0; j < 4; ++j) {
for (int k = 0; k < 2; ++k) {
char[] cs = now.toCharArray();
int digit = (cs[j] - '0' + downUp[k] + 10) % 10;
cs[j] = (char) ('0' + digit);
String nxt = new String(cs);
if (!vis.contains(nxt)) {
vis.add(nxt);
queue.offer(nxt);
}
}
}
}
turn ++;
}
return -1;
}不过我更喜欢数字版的，哈哈。
数字版本：
public int openLock(String[] deadends, String target) {
int tar = Integer.parseInt(target);
Set<Integer> vis = new HashSet<>();
for (int i = 0; i < deadends.length; ++i) {
vis.add(Integer.parseInt(deadends[i]));
}
Queue<Integer> queue = new ArrayDeque<>();
queue.offer(0);
if (vis.contains(0)) return -1;
vis.add(0);
int[] ten = {1000, 100, 10, 1};
int[] downUp = {1, -1};
int turn = 0;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; ++i) {
int now = queue.poll();
if (now == tar) {
return turn;
}
for (int j = 0; j < 4; ++j) {
for (int k = 0; k < 2; ++k) {
int[] digits = new int[4];
for (int l = 0; l < 4; ++l) {
digits[l] = now / ten[l] % 10;
}
digits[j] = (digits[j] + downUp[k] + 10) % 10;
int nxt = 0;
for (int l = 3; l >= 0; --l) {
nxt = 10 * nxt + digits[l];
}
if (!vis.contains(nxt)) {
vis.add(nxt);
queue.offer(nxt);
}
}
}
}
turn ++;
}
return -1;
}Python版本：
def openLock(self, deadends, target):
"""
:type deadends: List[str]
:type target: str
:rtype: int
"""
vis = set(deadends)
queue = []
queue.append("0000")
if ("0000" in vis): return -1
vis.add("0000")
turn = 0
while queue:
size = len(queue)
for i in range(size):
now = queue.pop(0)
if (now == target): return turn
for j in range(4):
for k in [-1, 1]:
digit = int(now[j])
digit = (digit + k + 10) % 10
nxt = now[: j] + str(digit) + now[j + 1 :]
if nxt not in vis:
vis.add(nxt)
queue.append(nxt)
turn += 1
return -1
