752. 打开转盘锁

题目描述

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，你需要给出解锁需要的最小旋转次数，如果无论如何不能解锁，返回 -1 。

示例 1:

输入：deadends = ["0201","0101","0102","1212","2002"], target = "0202"
输出：6
解释：
可能的移动序列为 "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202"。
注意 "0000" -> "0001" -> "0002" -> "0102" -> "0202" 这样的序列是不能解锁的，
因为当拨动到 "0102" 时这个锁就会被锁定。

示例 2:

输入: deadends = ["8888"], target = "0009"
输出：1
解释：把最后一位反向旋转一次即可 "0000" -> "0009"。

示例 3:

输入: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
输出：-1
解释：无法旋转到目标数字且不被锁定。

提示：

1 <= deadends.length <= 500
deadends[i].length == 4
target.length == 4
target 不在 deadends 之中
target 和 deadends[i] 仅由若干位数字组成

解法

方法一：朴素 BFS

直接用朴素 BFS。

class Solution:
    def openLock(self, deadends: List[str], target: str) -> int:
        def next(s):
            res = []
            s = list(s)
            for i in range(4):
                c = s[i]
                s[i] = '9' if c == '0' else str(int(c) - 1)
                res.append(''.join(s))
                s[i] = '0' if c == '9' else str(int(c) + 1)
                res.append(''.join(s))
                s[i] = c
            return res

        if target == '0000':
            return 0
        s = set(deadends)
        if '0000' in s:
            return -1
        q = deque([('0000')])
        s.add('0000')
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                p = q.popleft()
                for t in next(p):
                    if t == target:
                        return ans
                    if t not in s:
                        q.append(t)
                        s.add(t)
        return -1

class Solution {
    public int openLock(String[] deadends, String target) {
        if ("0000".equals(target)) {
            return 0;
        }
        Set<String> s = new HashSet<>(Arrays.asList(deadends));
        if (s.contains("0000")) {
            return -1;
        }
        Deque<String> q = new ArrayDeque<>();
        q.offer("0000");
        s.add("0000");
        int ans = 0;
        while (!q.isEmpty()) {
            ++ans;
            for (int n = q.size(); n > 0; --n) {
                String p = q.poll();
                for (String t : next(p)) {
                    if (target.equals(t)) {
                        return ans;
                    }
                    if (!s.contains(t)) {
                        q.offer(t);
                        s.add(t);
                    }
                }
            }
        }
        return -1;
    }

    private List<String> next(String t) {
        List res = new ArrayList<>();
        char[] chars = t.toCharArray();
        for (int i = 0; i < 4; ++i) {
            char c = chars[i];
            chars[i] = c == '0' ? '9' : (char) (c - 1);
            res.add(String.valueOf(chars));
            chars[i] = c == '9' ? '0' : (char) (c + 1);
            res.add(String.valueOf(chars));
            chars[i] = c;
        }
        return res;
    }
}

class Solution {
public:
    int openLock(vector<string>& deadends, string target) {
        unordered_set<string> s(deadends.begin(), deadends.end());
        if (s.count("0000")) return -1;
        if (target == "0000") return 0;
        queue<string> q{{"0000"}};
        s.insert("0000");
        int ans = 0;
        while (!q.empty()) {
            ++ans;
            for (int n = q.size(); n > 0; --n) {
                string p = q.front();
                q.pop();
                for (string t : next(p)) {
                    if (target == t) return ans;
                    if (!s.count(t)) {
                        q.push(t);
                        s.insert(t);
                    }
                }
            }
        }
        return -1;
    }

    vector<string> next(string& t) {
        vector<string> res;
        for (int i = 0; i < 4; ++i) {
            char c = t[i];
            t[i] = c == '0' ? '9' : (char) (c - 1);
            res.push_back(t);
            t[i] = c == '9' ? '0' : (char) (c + 1);
            res.push_back(t);
            t[i] = c;
        }
        return res;
    }
};

func openLock(deadends []string, target string) int {
	if target == "0000" {
		return 0
	}
	s := make(map[string]bool)
	for _, d := range deadends {
		s[d] = true
	}
	if s["0000"] {
		return -1
	}
	q := []string{"0000"}
	s["0000"] = true
	ans := 0
	next := func(t string) []string {
		s := []byte(t)
		var res []string
		for i, b := range s {
			s[i] = b - 1
			if s[i] < '0' {
				s[i] = '9'
			}
			res = append(res, string(s))
			s[i] = b + 1
			if s[i] > '9' {
				s[i] = '0'
			}
			res = append(res, string(s))
			s[i] = b
		}
		return res
	}
	for len(q) > 0 {
		ans++
		for n := len(q); n > 0; n-- {
			p := q[0]
			q = q[1:]
			for _, t := range next(p) {
				if target == t {
					return ans
				}
				if !s[t] {
					q = append(q, t)
					s[t] = true
				}
			}
		}
	}
	return -1
}

方法二：双向 BFS

本题也可以用双向 BFS 优化搜索空间，从而提升效率。

双向 BFS 是 BFS 常见的一个优化方法，主要实现思路如下：

创建两个队列 q1, q2 分别用于“起点 -> 终点”、“终点 -> 起点”两个方向的搜索；
创建两个哈希表 m1, m2 分别记录访问过的节点以及对应的扩展次数（步数）；
每次搜索时，优先选择元素数量较少的队列进行搜索扩展，如果在扩展过程中，搜索到另一个方向已经访问过的节点，说明找到了最短路径；
只要其中一个队列为空，说明当前方向的搜索已经进行不下去了，说明起点到终点不连通，无需继续搜索。

while q1 and q2:
    if len(q1) <= len(q2):
        # 优先选择较少元素的队列进行扩展
        extend(m1, m2, q1)
    else:
        extend(m2, m1, q2)


def extend(m1, m2, q):
    # 新一轮扩展
    for _ in range(len(q)):
        p = q.popleft()
        step = m1[p]
        for t in next(p):
            if t in m1:
                # 此前已经访问过
                continue
            if t in m2:
                # 另一个方向已经搜索过，说明找到了一条最短的连通路径
                return step + 1 + m2[t]
            q.append(t)
            m1[t] = step + 1

class Solution:
    def openLock(self, deadends: List[str], target: str) -> int:
        def next(s):
            res = []
            s = list(s)
            for i in range(4):
                c = s[i]
                s[i] = '9' if c == '0' else str(int(c) - 1)
                res.append(''.join(s))
                s[i] = '0' if c == '9' else str(int(c) + 1)
                res.append(''.join(s))
                s[i] = c
            return res

        def extend(m1, m2, q):
            for _ in range(len(q)):
                p = q.popleft()
                step = m1[p]
                for t in next(p):
                    if t in s or t in m1:
                        continue
                    if t in m2:
                        return step + 1 + m2[t]
                    m1[t] = step + 1
                    q.append(t)
            return -1

        def bfs():
            m1, m2 = {"0000": 0}, {target: 0}
            q1, q2 = deque([('0000')]), deque([(target)])
            while q1 and q2:
                t = extend(m1, m2, q1) if len(q1) <= len(q2) else extend(m2, m1, q2)
                if t != -1:
                    return t
            return -1

        if target == '0000':
            return 0
        s = set(deadends)
        if '0000' in s:
            return -1
        return bfs()

class Solution {
    private String start;
    private String target;
    private Set<String> s = new HashSet<>();

    public int openLock(String[] deadends, String target) {
        if ("0000".equals(target)) {
            return 0;
        }
        start = "0000";
        this.target = target;
        for (String d : deadends) {
            s.add(d);
        }
        if (s.contains(start)) {
            return -1;
        }
        return bfs();
    }

    private int bfs() {
        Map<String, Integer> m1 = new HashMap<>();
        Map<String, Integer> m2 = new HashMap<>();
        Deque<String> q1 = new ArrayDeque<>();
        Deque<String> q2 = new ArrayDeque<>();
        m1.put(start, 0);
        m2.put(target, 0);
        q1.offer(start);
        q2.offer(target);
        while (!q1.isEmpty() && !q2.isEmpty()) {
            int t = q1.size() <= q2.size() ? extend(m1, m2, q1) : extend(m2, m1, q2);
            if (t != -1) {
                return t;
            }
        }
        return -1;
    }

    private int extend(Map<String, Integer> m1, Map<String, Integer> m2, Deque<String> q) {
        for (int n = q.size(); n > 0; --n) {
            String p = q.poll();
            int step = m1.get(p);
            for (String t : next(p)) {
                if (m1.containsKey(t) || s.contains(t)) {
                    continue;
                }
                if (m2.containsKey(t)) {
                    return step + 1 + m2.get(t);
                }
                m1.put(t, step + 1);
                q.offer(t);
            }
        }
        return -1;
    }

    private List<String> next(String t) {
        List res = new ArrayList<>();
        char[] chars = t.toCharArray();
        for (int i = 0; i < 4; ++i) {
            char c = chars[i];
            chars[i] = c == '0' ? '9' : (char) (c - 1);
            res.add(String.valueOf(chars));
            chars[i] = c == '9' ? '0' : (char) (c + 1);
            res.add(String.valueOf(chars));
            chars[i] = c;
        }
        return res;
    }
}

class Solution {
public:
    unordered_set<string> s;
    string start;
    string target;

    int openLock(vector<string>& deadends, string target) {
        if (target == "0000") return 0;
        for (auto d : deadends) s.insert(d);
        if (s.count("0000")) return -1;
        this->start = "0000";
        this->target = target;
        return bfs();
    }

    int bfs() {
        unordered_map<string, int> m1;
        unordered_map<string, int> m2;
        m1[start] = 0;
        m2[target] = 0;
        queue<string> q1{{start}};
        queue<string> q2{{target}};
        while (!q1.empty() && !q2.empty()) {
            int t = q1.size() <= q2.size() ? extend(m1, m2, q1) : extend(m2, m1, q2);
            if (t != -1) return t;
        }
        return -1;
    }

    int extend(unordered_map<string, int>& m1, unordered_map<string, int>& m2, queue<string>& q) {
        for (int n = q.size(); n > 0; --n) {
            string p = q.front();
            int step = m1[p];
            q.pop();
            for (string t : next(p)) {
                if (s.count(t) || m1.count(t)) continue;
                if (m2.count(t)) return step + 1 + m2[t];
                m1[t] = step + 1;
                q.push(t);
            }
        }
        return -1;
    }

    vector<string> next(string& t) {
        vector<string> res;
        for (int i = 0; i < 4; ++i) {
            char c = t[i];
            t[i] = c == '0' ? '9' : (char) (c - 1);
            res.push_back(t);
            t[i] = c == '9' ? '0' : (char) (c + 1);
            res.push_back(t);
            t[i] = c;
        }
        return res;
    }
};

func openLock(deadends []string, target string) int {
	if target == "0000" {
		return 0
	}
	s := make(map[string]bool)
	for _, d := range deadends {
		s[d] = true
	}
	if s["0000"] {
		return -1
	}
	next := func(t string) []string {
		s := []byte(t)
		var res []string
		for i, b := range s {
			s[i] = b - 1
			if s[i] < '0' {
				s[i] = '9'
			}
			res = append(res, string(s))
			s[i] = b + 1
			if s[i] > '9' {
				s[i] = '0'
			}
			res = append(res, string(s))
			s[i] = b
		}
		return res
	}

	extend := func(m1, m2 map[string]int, q *[]string) int {
		for n := len(*q); n > 0; n-- {
			p := (*q)[0]
			*q = (*q)[1:]
			step, _ := m1[p]
			for _, t := range next(p) {
				if s[t] {
					continue
				}
				if _, ok := m1[t]; ok {
					continue
				}
				if v, ok := m2[t]; ok {
					return step + 1 + v
				}
				m1[t] = step + 1
				*q = append(*q, t)
			}
		}
		return -1
	}

	bfs := func() int {
		q1 := []string{"0000"}
		q2 := []string{target}
		m1 := map[string]int{"0000": 0}
		m2 := map[string]int{target: 0}
		for len(q1) > 0 && len(q2) > 0 {
			t := -1
			if len(q1) <= len(q2) {
				t = extend(m1, m2, &q1)
			} else {
				t = extend(m2, m1, &q2)
			}
			if t != -1 {
				return t
			}
		}
		return -1
	}

	return bfs()
}

方法三：A*算法

A* 算法主要思想如下：

将 BFS 队列转换为优先队列（小根堆）；
队列中的每个元素为 (dist[state] + f(state), state)，dist[state] 表示从起点到当前 state 的距离，f(state) 表示从当前 state 到终点的估计距离，这两个距离之和作为堆排序的依据；
当终点第一次出队时，说明找到了从起点到终点的最短路径，直接返回对应的 step；
f(state) 是估价函数，并且估价函数要满足 f(state) <= g(state)，其中 g(state) 表示 state 到终点的真实距离；
A* 算法只能保证终点第一次出队时，即找到了一条从起点到终点的最小路径，不能保证其他点出队时也是从起点到当前点的最短路径。

class Solution:
    def openLock(self, deadends: List[str], target: str) -> int:
        def next(s):
            res = []
            s = list(s)
            for i in range(4):
                c = s[i]
                s[i] = '9' if c == '0' else str(int(c) - 1)
                res.append(''.join(s))
                s[i] = '0' if c == '9' else str(int(c) + 1)
                res.append(''.join(s))
                s[i] = c
            return res

        def f(s):
            ans = 0
            for i in range(4):
                a = ord(s[i]) - ord('0')
                b = ord(target[i]) - ord('0')
                if a > b:
                    a, b = b, a
                ans += min(b - a, a + 10 - b)
            return ans

        if target == '0000':
            return 0
        s = set(deadends)
        if '0000' in s:
            return -1
        start = '0000'
        q = [(f(start), start)]
        dist = {start: 0}
        while q:
            _, state = heappop(q)
            if state == target:
                return dist[state]
            for t in next(state):
                if t in s:
                    continue
                if t not in dist or dist[t] > dist[state] + 1:
                    dist[t] = dist[state] + 1
                    heappush(q, (dist[t] + f(t), t))
        return -1

class Solution {
    private String target;

    public int openLock(String[] deadends, String target) {
        if ("0000".equals(target)) {
            return 0;
        }
        String start = "0000";
        this.target = target;
        Set<String> s = new HashSet<>();
        for (String d : deadends) {
            s.add(d);
        }
        if (s.contains(start)) {
            return -1;
        }
        PriorityQueue<Pair<Integer, String>> q
            = new PriorityQueue<>(Comparator.comparingInt(Pair::getKey));
        q.offer(new Pair<>(f(start), start));
        Map<String, Integer> dist = new HashMap<>();
        dist.put(start, 0);
        while (!q.isEmpty()) {
            String state = q.poll().getValue();
            int step = dist.get(state);
            if (target.equals(state)) {
                return step;
            }
            for (String t : next(state)) {
                if (s.contains(t)) {
                    continue;
                }
                if (!dist.containsKey(t) || dist.get(t) > step + 1) {
                    dist.put(t, step + 1);
                    q.offer(new Pair<>(step + 1 + f(t), t));
                }
            }
        }
        return -1;
    }

    private int f(String s) {
        int ans = 0;
        for (int i = 0; i < 4; ++i) {
            int a = s.charAt(i) - '0';
            int b = target.charAt(i) - '0';
            if (a > b) {
                int t = a;
                a = b;
                b = a;
            }
            ans += Math.min(b - a, a + 10 - b);
        }
        return ans;
    }

    private List<String> next(String t) {
        List res = new ArrayList<>();
        char[] chars = t.toCharArray();
        for (int i = 0; i < 4; ++i) {
            char c = chars[i];
            chars[i] = c == '0' ? '9' : (char) (c - 1);
            res.add(String.valueOf(chars));
            chars[i] = c == '9' ? '0' : (char) (c + 1);
            res.add(String.valueOf(chars));
            chars[i] = c;
        }
        return res;
    }
}

class Solution {
public:
    string target;

    int openLock(vector<string>& deadends, string target) {
        if (target == "0000") return 0;
        unordered_set<string> s(deadends.begin(), deadends.end());
        if (s.count("0000")) return -1;
        string start = "0000";
        this->target = target;
        typedef pair<int, string> PIS;
        priority_queue<PIS, vector<PIS>, greater<PIS>> q;
        unordered_map<string, int> dist;
        dist[start] = 0;
        q.push({f(start), start});
        while (!q.empty()) {
            PIS t = q.top();
            q.pop();
            string state = t.second;
            int step = dist[state];
            if (state == target) return step;
            for (string& t : next(state)) {
                if (s.count(t)) continue;
                if (!dist.count(t) || dist[t] > step + 1) {
                    dist[t] = step + 1;
                    q.push({step + 1 + f(t), t});
                }
            }
        }
        return -1;
    }

    int f(string s) {
        int ans = 0;
        for (int i = 0; i < 4; ++i) {
            int a = s[i] - '0';
            int b = target[i] - '0';
            if (a < b) {
                int t = a;
                a = b;
                b = t;
            }
            ans += min(b - a, a + 10 - b);
        }
        return ans;
    }

    vector<string> next(string& t) {
        vector<string> res;
        for (int i = 0; i < 4; ++i) {
            char c = t[i];
            t[i] = c == '0' ? '9' : (char) (c - 1);
            res.push_back(t);
            t[i] = c == '9' ? '0' : (char) (c + 1);
            res.push_back(t);
            t[i] = c;
        }
        return res;
    }
};

peaking / leetcode

752. 打开转盘锁

题目描述

解法

方法一：朴素 BFS

方法二：双向 BFS

方法三：A*算法

简介

发行版

贡献者

近期动态

peaking / leetcode .gitee-modal { width: 500px !important; }

752. 打开转盘锁

题目描述

解法

方法一：朴素 BFS

方法二：双向 BFS

方法三：A*算法

简介

发行版

贡献者

近期动态

搜索帮助

peaking / leetcode