代码拉取完成,页面将自动刷新
同步操作将从 doocs/leetcode 强制同步,此操作会覆盖自 Fork 仓库以来所做的任何修改,且无法恢复!!!
确定后同步将在后台操作,完成时将刷新页面,请耐心等待。
Given an array of digits
which is sorted in non-decreasing order. You can write numbers using each digits[i]
as many times as we want. For example, if digits = ['1','3','5']
, we may write numbers such as '13'
, '551'
, and '1351315'
.
Return the number of positive integers that can be generated that are less than or equal to a given integer n
.
Example 1:
Input: digits = ["1","3","5","7"], n = 100 Output: 20 Explanation: The 20 numbers that can be written are: 1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.
Example 2:
Input: digits = ["1","4","9"], n = 1000000000 Output: 29523 Explanation: We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers, 81 four digit numbers, 243 five digit numbers, 729 six digit numbers, 2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers. In total, this is 29523 integers that can be written using the digits array.
Example 3:
Input: digits = ["7"], n = 8 Output: 1
Constraints:
1 <= digits.length <= 9
digits[i].length == 1
digits[i]
is a digit from '1'
to '9'
.digits
are unique.digits
is sorted in non-decreasing order.1 <= n <= 109
class Solution:
def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:
@cache
def dfs(pos, lead, limit):
if pos <= 0:
return lead == False
up = a[pos] if limit else 9
ans = 0
for i in range(up + 1):
if i == 0 and lead:
ans += dfs(pos - 1, lead, limit and i == up)
elif i in s:
ans += dfs(pos - 1, False, limit and i == up)
return ans
l = 0
a = [0] * 12
s = {int(d) for d in digits}
while n:
l += 1
a[l] = n % 10
n //= 10
return dfs(l, True, True)
class Solution {
private int[] a = new int[12];
private int[][] dp = new int[12][2];
private Set<Integer> s = new HashSet<>();
public int atMostNGivenDigitSet(String[] digits, int n) {
for (var e : dp) {
Arrays.fill(e, -1);
}
for (String d : digits) {
s.add(Integer.parseInt(d));
}
int len = 0;
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 1, true);
}
private int dfs(int pos, int lead, boolean limit) {
if (pos <= 0) {
return lead ^ 1;
}
if (!limit && lead != 1 && dp[pos][lead] != -1) {
return dp[pos][lead];
}
int ans = 0;
int up = limit ? a[pos] : 9;
for (int i = 0; i <= up; ++i) {
if (i == 0 && lead == 1) {
ans += dfs(pos - 1, lead, limit && i == up);
} else if (s.contains(i)) {
ans += dfs(pos - 1, 0, limit && i == up);
}
}
if (!limit && lead == 0) {
dp[pos][lead] = ans;
}
return ans;
}
}
class Solution {
public:
int a[12];
int dp[12][2];
unordered_set<int> s;
int atMostNGivenDigitSet(vector<string>& digits, int n) {
memset(dp, -1, sizeof dp);
for (auto& d : digits) {
s.insert(stoi(d));
}
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 1, true);
}
int dfs(int pos, int lead, bool limit) {
if (pos <= 0) {
return lead ^ 1;
}
if (!limit && !lead && dp[pos][lead] != -1) {
return dp[pos][lead];
}
int ans = 0;
int up = limit ? a[pos] : 9;
for (int i = 0; i <= up; ++i) {
if (i == 0 && lead) {
ans += dfs(pos - 1, lead, limit && i == up);
} else if (s.count(i)) {
ans += dfs(pos - 1, 0, limit && i == up);
}
}
if (!limit && !lead) {
dp[pos][lead] = ans;
}
return ans;
}
};
func atMostNGivenDigitSet(digits []string, n int) int {
s := map[int]bool{}
for _, d := range digits {
i, _ := strconv.Atoi(d)
s[i] = true
}
a := make([]int, 12)
dp := make([][2]int, 12)
for i := range a {
dp[i] = [2]int{-1, -1}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}
var dfs func(int, int, bool) int
dfs = func(pos, lead int, limit bool) int {
if pos <= 0 {
return lead ^ 1
}
if !limit && lead == 0 && dp[pos][lead] != -1 {
return dp[pos][lead]
}
up := 9
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
if i == 0 && lead == 1 {
ans += dfs(pos-1, lead, limit && i == up)
} else if s[i] {
ans += dfs(pos-1, 0, limit && i == up)
}
}
if !limit {
dp[pos][lead] = ans
}
return ans
}
return dfs(l, 1, true)
}
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。