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1707 · Knight Dialer
Algorithms
Medium
Accepted Rate44%
Description
Solution
Notes
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Description
A chess knight can move as indicated in the chess diagram below:
This time, we place our chess knight on any numbered key of a phone pad (indicated above), and the knight makes N-1 hops. Each hop must be from one key to another numbered key.
Each time it lands on a key (including the initial placement of the knight), it presses the number of that key, pressing N digits total.
How many distinct numbers can you dial in this manner?
Since the answer may be large, output the answer modulo 10^9 + 7.
1 \leq N \leq 50001≤N≤5000
Example
Example 1:
Input: 1 Output: 10 Explanation: The answer may be 0, 1, 2, 3, ..., 9. Example 2:
Input: 2 Output: 20 Explanation: The answer may be 04, 06, 16, 18, 27, 29, 34, 38, 43, 49, 40, 61, 67, 60, 72, 76, 81, 83, 94, 92. Example 3:
Input: 3 Output: 46 Tags
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解法1：DP。 dp[i][j]表示第i次跳到数字j时的distinct number数。 注意:
- 1e+9之类的科学计数法默认是double。但如果我们定义成int M = 1e+9则可将其强转为int。注意1e+9仍在int范围内。
- dp[1][i]初始化为1, 不是i。因为每个数字只能算一种情况。
- unordered_set可以直接用vector<vector>代替，因为序号0..9都是按顺序来的，可以当数组的索引。
class Solution {
public:
/**
* @param N: N
* @return: return the number of distinct numbers can you dial in this manner mod 1e9+7
*/
int knightDialer(int N) {
//#define M (long long)(10e9 + 7)
int M = 1e9 + 7;
unordered_map<int, vector<int>> dict =
{ {0, {4, 6}},
{1, {6, 8}},
{2, {7, 9}},
{3, {4, 8}},
{4, {0, 3, 9}},
{5, {}},
{6, {0, 1, 7}},
{7, {2, 6}},
{8, {1, 3}},
{9, {2, 4}}
};
vector<vector<int>> dp(N + 1, vector<int>(10, 0));
for (int i = 0; i <= 9; i++) {
dp[1][i] = 1;
}
for (int i = 2; i <= N; i++) {
for (int j = 0; j <= 9; j++) {
for (int k = 0; k < dict[j].size(); k++) {
dp[i][j] = (dp[i][j] + dp[i - 1][dict[j][k]]) % M;
}
}
}
int sum = 0;
for (int i = 0; i <= 9; ++i) {
sum = (sum + dp[N][i]) % M;
}
return sum;
}
};
