351. Android Unlock Patterns #

题目 #

Android devices have a special lock screen with a 3 x 3 grid of dots. Users can set an “unlock pattern” by connecting the dots in a specific sequence, forming a series of joined line segments where each segment’s endpoints are two consecutive dots in the sequence. A sequence of k dots is a valid unlock pattern if both of the following are true:

• All the dots in the sequence are distinct.
• If the line segment connecting two consecutive dots in the sequence passes through the center of any other dot, the other dot must have previously appeared in the sequence. No jumps through non-selected dots are allowed.

Given two integers m and n, return the number of unique and valid unlock patterns of the Android grid lock screen that consist of at least m keys and at most n keys.

Two unlock patterns are considered unique if there is a dot in one sequence that is not in the other, or the order of the dots is different.

Example 1:

Input: m = 1, n = 1
Output: 9

Example 2:

Input: m = 1, n = 2
Output: 65

Constraints:

• 1 <= m, n <= 9

解题思路 #

• 回溯。3x3 网格上从某个点滑到另一个点时，如果中间正好跨过第三个点（如 1->3 跨过 2，1->9 跨过 5），那么中间这个点必须已经被选过，否则该步非法。
• 用 blocker[a][b] 记录从 a 到 b 必须先经过的中间点，0 表示相邻无需跨点。设置一个虚拟起点 0，从它出发可以到任意数字，然后 DFS 统计长度落在 [m, n] 内的所有合法路径数。

代码 #

package leetcode

// blocker351[a][b] 记录从点 a 滑到点 b 必须先经过的中间点；0 表示两点相邻、无需经过中间点。
// 例如 1->3 必须先经过 2，1->9 必须先经过 5。
var blocker351 = [10][10]int{
	1: {3: 2, 7: 4, 9: 5},
	2: {8: 5},
	3: {1: 2, 7: 5, 9: 6},
	4: {6: 5},
	6: {4: 5},
	7: {1: 4, 3: 5, 9: 8},
	8: {2: 5},
	9: {1: 5, 3: 6, 7: 8},
}

// 解法一 回溯。0 号点是虚拟起点，从它出发可以到任意一个数字。
func numberOfPatterns(m int, n int) int {
	visited := make([]bool, 10)
	return countPatterns351(visited, 0, 0, m, n)
}

// now 是当前所在的点，already 是已经选中的点数，统计长度落在 [m,n] 内的合法图案数。
func countPatterns351(visited []bool, now, already, m, n int) int {
	res := 0
	if already >= m && already <= n {
		res++
	}
	if already >= n { // 已达到最大长度，无需再往下选
		return res
	}
	for next := 1; next <= 9; next++ {
		mid := blocker351[now][next]
		// next 没被用过，且 next 与 now 相邻（mid==0）或中间点已被划过（visited[mid]）
		if !visited[next] && (mid == 0 || visited[mid]) {
			visited[next] = true
			res += countPatterns351(visited, next, already+1, m, n)
			visited[next] = false
		}
	}
	return res
}