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。
A message containing letters from A-Z can be encoded into numbers using the following mapping:
'A' -> "1" 'B' -> "2" ... 'Z' -> "26"
To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:
"AAJF" with the grouping (1 1 10 6)
"KJF" with the grouping (11 10 6)
Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".
Given a string s containing only digits, return the number of ways to decode it.
The answer is guaranteed to fit in a 32-bit integer.
Example 1:
Input: s = "12" Output: 2 Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: s = "226" Output: 3 Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
Example 3:
Input: s = "0" Output: 0 Explanation: There is no character that is mapped to a number starting with 0. The only valid mappings with 0 are 'J' -> "10" and 'T' -> "20", neither of which start with 0. Hence, there are no valid ways to decode this since all digits need to be mapped.
Example 4:
Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").
Constraints:
1 <= s.length <= 100s contains only digits and may contain leading zero(s).Solution1:
class Solution:
def numDecodings(self, s: str) -> int:
n = len(s)
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n + 1):
if s[i - 1] != '0':
dp[i] += dp[i - 1]
if i > 1 and s[i - 2] != '0' and (int(s[i - 2]) * 10 + int(s[i - 1]) <= 26):
dp[i] += dp[i - 2]
return dp[n]
Solution2:
class Solution:
def numDecodings(self, s: str) -> int:
n = len(s)
a, b, c = 0, 1, 0
for i in range(1, n + 1):
c = 0
if s[i - 1] != '0':
c += b
if i > 1 and s[i - 2] != '0' and (int(s[i - 2]) * 10 + int(s[i - 1]) <= 26):
c += a
a, b = b, c
return c
Solution1:
class Solution {
public int numDecodings(String s) {
int n = s.length();
int[] dp = new int[n + 1];
dp[0] = 1;
for (int i = 1; i <= n; ++i) {
if (s.charAt(i - 1) != '0') {
dp[i] += dp[i - 1];
}
if (i > 1 && s.charAt(i - 2) != '0' && ((s.charAt(i - 2) - '0') * 10 + s.charAt(i - 1) - '0') <= 26) {
dp[i] += dp[i - 2];
}
}
return dp[n];
}
}
Solution2:
class Solution {
public int numDecodings(String s) {
int n = s.length();
int a = 0, b = 1, c = 0;
for (int i = 1; i <= n; ++i) {
c = 0;
if (s.charAt(i - 1) != '0') {
c += b;
}
if (i > 1 && s.charAt(i - 2) != '0' && ((s.charAt(i - 2) - '0') * 10 + s.charAt(i - 1) - '0') <= 26) {
c += a;
}
a = b;
b = c;
}
return c;
}
}
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。