Leetcode: Android(安卓)Unlock Patterns

Given an Android 3x3 key lock screen and two integers m and n, where 1 ≤ m ≤ n ≤ 9, count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys.Rules for a valid pattern:Each pattern must connect at least m keys and at most n keys.All the keys must be distinct.If the line connecting two consecutive keys in the pattern passes through any other keys, the other keys must have previously selected in the pattern. No jumps through non selected key is allowed.The order of keys used matters.Explanation:| 1 | 2 | 3 || 4 | 5 | 6 || 7 | 8 | 9 |Invalid move: 4 - 1 - 3 - 6 Line 1 - 3 passes through key 2 which had not been selected in the pattern.Invalid move: 4 - 1 - 9 - 2Line 1 - 9 passes through key 5 which had not been selected in the pattern.Valid move: 2 - 4 - 1 - 3 - 6Line 1 - 3 is valid because it passes through key 2, which had been selected in the patternValid move: 6 - 5 - 4 - 1 - 9 - 2Line 1 - 9 is valid because it passes through key 5, which had been selected in the pattern.Example:Given m = 1, n = 1, return 9.

我自己的backtracking做法

最开始把cur设置为一个dummy value 0

 1 public class Solution { 2     int num = 0; 3     public int numberOfPatterns(int m, int n) { 4         for (int len=m; len<=n; len++) { 5             HashSet<Integer> visited = new HashSet<Integer>(); 6             count(visited, 0, 0, len); 7         } 8         return num; 9     }10     11     public void count(HashSet<Integer> visited, int cur, int pos, int len) {12         if (pos == len) {13             num++;14             return;15         }16         for (int elem=1; elem<=9; elem++) {17             if (visited.contains(elem)) continue;18             if (cur == 1) {19                 if (elem==3 && !visited.contains(2)) continue;20                 if (elem==7 && !visited.contains(4)) continue;21                 if (elem==9 && !visited.contains(5)) continue;22             }23             else if (cur == 2) {24                 if (elem == 8 && !visited.contains(5)) continue;25             }26             else if (cur == 3) {27                 if (elem==1 && !visited.contains(2)) continue;28                 if (elem==7 && !visited.contains(5)) continue;29                 if (elem==9 && !visited.contains(6)) continue;30             }31             else if (cur == 4) {32                 if (elem == 6 && !visited.contains(5)) continue;33             }34             else if (cur == 6) {35                 if (elem == 4 && !visited.contains(5)) continue;36             }37             else if (cur == 7) {38                 if (elem==1 && !visited.contains(4)) continue;39                 if (elem==3 && !visited.contains(5)) continue;40                 if (elem==9 && !visited.contains(8)) continue;41             }42             else if (cur == 8) {43                 if (elem==2 && !visited.contains(5)) continue;44             }45             else if (cur == 9) {46                 if (elem==1 && !visited.contains(5)) continue;47                 if (elem==3 && !visited.contains(6)) continue;48                 if (elem==7 && !visited.contains(8)) continue;49             }50             visited.add(elem);51             count(visited, elem, pos+1, len);52             visited.remove(elem);53         }54     }55 }

最好的DFS with Optimization beat 97%: https://discuss.leetcode.com/topic/46260/java-dfs-solution-with-clear-explanations-and-optimization-beats-97-61-12ms

Use an matrix to store the corssed number for each possible move and use DFS to find out all patterns.

The optimization idea is that 1,3,7,9 are symmetric, 2,4,6,8 are also symmetric. Hence we only calculate one among each group and multiply by 4.

 1 public class Solution { 2     // cur: the current position 3     // remain: the steps remaining 4     int DFS(boolean vis[], int[][] skip, int cur, int remain) { 5         if(remain < 0) return 0; 6         if(remain == 0) return 1; 7         vis[cur] = true; 8         int rst = 0; 9         for(int i = 1; i <= 9; ++i) {10             // If vis[i] is not visited and (two numbers are adjacent or skip number is already visited)11             if(!vis[i] && (skip[cur][i] == 0 || (vis[skip[cur][i]]))) {12                 rst += DFS(vis, skip, i, remain - 1);13             }14         }15         vis[cur] = false;16         return rst;17     }18     19     public int numberOfPatterns(int m, int n) {20         // Skip array represents number to skip between two pairs21         int skip[][] = new int[10][10];22         skip[1][3] = skip[3][1] = 2;23         skip[1][7] = skip[7][1] = 4;24         skip[3][9] = skip[9][3] = 6;25         skip[7][9] = skip[9][7] = 8;26         skip[1][9] = skip[9][1] = skip[2][8] = skip[8][2] = skip[3][7] = skip[7][3] = skip[4][6] = skip[6][4] = 5;27         boolean vis[] = new boolean[10];28         int rst = 0;29         // DFS search each length from m to n30         for(int i = m; i <= n; ++i) {31             rst += DFS(vis, skip, 1, i - 1) * 4;    // 1, 3, 7, 9 are symmetric32             rst += DFS(vis, skip, 2, i - 1) * 4;    // 2, 4, 6, 8 are symmetric33             rst += DFS(vis, skip, 5, i - 1);        // 534         }35         return rst;36     }37 }

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