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:
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".
Input: deadends = ["8888"], target = "0009"
Output: 1
Explanation:
We can turn the last wheel in reverse to move from "0000" -> "0009".
Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
Output: -1
Explanation:
We can't reach the target without getting stuck.
Input: deadends = ["0000"], target = "8888"
Output: -1
我们可以将 0000 到 9999 这 10000 状态看成图上的 10000 个节点，两个节点之间存在一条边，当且仅当这两个节点对应的状态只有 1 位不同，且不同的那位相差 1（包括 0 和 9 也相差 1 的情况），并且这两个节点均不在数组 deadends 中。最终的答案即为 0000 到 target 的最短路径。
我们用广度优先搜索来找到最短路径，从 0000 开始搜索。对于每一个状态，它可以扩展到最多 8 个状态，即将它的第 i = 0, 1, 2, 3 位增加 1 或减少 1，将这些状态中没有搜索过并且不在 deadends 中的状态全部加入到队列中，并继续进行搜索。注意 0000 本身有可能也在 deadends 中。
class Solution(object):
def openLock(self, deadends, target):
def neighbors(node):
for i in range(4):
x = int(node[i])
for d in (-1, 1):
y = (x + d) % 10
yield node[:i] + str(y) + node[i+1:]
dead = set(deadends)
queue = collections.deque([('0000', 0)])
seen = {'0000'}
while queue:
node, depth = queue.popleft()
if node == target: return depth
if node in dead: continue
for nei in neighbors(node):
if nei not in seen:
seen.add(nei)
queue.append((nei, depth+1))
return -1执行用时：664 ms, 在所有 Python3 提交中击败了51.34%的用户
内存消耗：16.1 MB, 在所有 Python3 提交中击败了42.43%的用户
Attention:
- 表盘问题使用广度优先搜索
- yield in python
- node, depth = queue.popleft()
c++
class Solution {
public:
std::vector<string> neighbors(string& cur){
std::vector<string> neighbors;
for (int i = 0; i < 4; i++){
for (const int& d : {-1, 1}){
string nei_left = cur.substr(0, i);
string nei_right = cur.substr(i+1, 3-i);
int nei_mid = (cur[i] - '0' + d) % 10;
if (nei_mid < 0) nei_mid += 10;
string neighbor = nei_left + to_string(nei_mid) + nei_right;
neighbors.push_back(neighbor);
}
}
return neighbors;
}
int openLock(std::vector<string>& deadends, string target) {
std::set<string> seen = {"0000"};
std::queue<std::pair<string, int>> Q;
Q.push(std::pair<string, int>("0000", 0));
while (!Q.empty()){
string cur = Q.front().first;
int step = Q.front().second;
Q.pop();
if (cur == target)
return step;
if (find(deadends.begin(), deadends.end(), cur) != deadends.end())
continue;
std::vector<string> neis = neighbors(cur);
for (const string& nei : neis){
if (seen.find(nei) == seen.end()){
seen.insert(nei);
Q.push(std::pair<string, int>(nei, step + 1));
}
}
}
return -1;
}
};执行用时：1064 ms, 在所有 C++ 提交中击败了5.11%的用户
内存消耗：138.7 MB, 在所有 C++ 提交中击败了5.05%的用户
Attention:
- int(char) 会转为ascii码，正确做法是 ch - '0'
- c++求余会求出符数，如果小于0，需要加余数