752. Open the Lock

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.

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

Constraints:

1 <= deadends.length <= 500
deadends[i].length == 4
target.length == 4
target will not be in the list deadends.
target and deadends[i] consist of digits only.

题目大意

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 。

解题思路

此题可以转化为从起始点到终点的最短路径。采用广度优先搜索。每次广搜枚举转动一次数字的状态，并且用 visited 记录是否被搜索过，如果没有被搜索过，便加入队列，下一轮继续搜索。如果搜索到了 target，便返回对应的旋转次数。如果搜索完成后，仍没有搜索到 target，说明无法解锁，返回 -1。特殊情况，如果 target 就是初始数字 0000，那么直接返回答案 0。
在广搜之前，先将 deadends 放入 map 中，搜索中判断是否搜到了 deadends。如果初始数字 0000 出现在 deadends 中，可以直接返回答案 −1。

代码

package leetcode

func openLock(deadends []string, target string) int {
	if target == "0000" {
		return 0
	}
	targetNum, visited := strToInt(target), make([]bool, 10000)
	visited[0] = true
	for _, deadend := range deadends {
		num := strToInt(deadend)
		if num == 0 {
			return -1
		}
		visited[num] = true
	}
	depth, curDepth, nextDepth := 0, []int16{0}, make([]int16, 0)
	var nextNum int16
	for len(curDepth) > 0 {
		nextDepth = nextDepth[0:0]
		for _, curNum := range curDepth {
			for incrementer := int16(1000); incrementer > 0; incrementer /= 10 {
				digit := (curNum / incrementer) % 10
				if digit == 9 {
					nextNum = curNum - 9*incrementer
				} else {
					nextNum = curNum + incrementer
				}
				if nextNum == targetNum {
					return depth + 1
				}
				if !visited[nextNum] {
					visited[nextNum] = true
					nextDepth = append(nextDepth, nextNum)
				}
				if digit == 0 {
					nextNum = curNum + 9*incrementer
				} else {
					nextNum = curNum - incrementer
				}
				if nextNum == targetNum {
					return depth + 1
				}
				if !visited[nextNum] {
					visited[nextNum] = true
					nextDepth = append(nextDepth, nextNum)
				}
			}
		}
		curDepth, nextDepth = nextDepth, curDepth
		depth++
	}
	return -1
}

func strToInt(str string) int16 {
	return int16(str[0]-'0')*1000 + int16(str[1]-'0')*100 + int16(str[2]-'0')*10 + int16(str[3]-'0')
}

Uh oh!