467 Unique Substrings in Wraparound String 环绕字符串中唯一的子字符串
Description:
Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
Now we have another string p. Your job is to find out how many unique non-empty substrings of p are present in s. In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.
Note:
p consists of only lowercase English letters and the size of p might be over 10000.
Example:
Example 1:
Input: "a"
Output: 1
Explanation: Only the substring "a" of string "a" is in the string �s.
Example 2:
Input: "cac"
Output: 2
Explanation: There are two substrings "a", "c" of string "cac" in the string s.
Example 3:
Input: "zab"
Output: 6
Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.
题目描述:
把字符串 s 看作是“abcdefghijklmnopqrstuvwxyz”的无限环绕字符串，所以 s 看起来是这样的："...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...."
现在我们有了另一个字符串 p 。你需要的是找出 s 中有多少个唯一的 p 的非空子串，尤其是当你的输入是字符串 p ，你需要输出字符串 s 中 p 的不同的非空子串的数目。
注意:
p 仅由小写的英文字母组成，p 的大小可能超过 10000。
示例 :
示例 1:
输入: "a"
输出: 1
解释: 字符串 S 中只有一个"a"子字符。
示例 2:
输入: "cac"
输出: 2
解释: 字符串 S 中的字符串“cac”只有两个子串“a”、“c”。
示例 3:
输入: "zab"
输出: 6
解释: 在字符串 S 中有六个子串“z”、“a”、“b”、“za”、“ab”、“zab”。
思路:
动态规划
统计以各字母结尾的最长连续递增子字符串的长度, 再求和
比如 abcdbcd, a -> 1, b -> 2(ab), c -> 3(abc, abcdbc不行因为不是连续递增的)
只看 abc, 有 a, b, c, ab, bc, abc共 6个刚好是前面的和
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int findSubstringInWraproundString(string p)
{
int n = p.size(), cur = 1;
vector count(26, 0);
for (int i = 0; i < n; i++)
{
if (i and (p[i] - p[i - 1] == 1 or p[i - 1] - p[i] == 25)) ++cur;
else cur = 1;
count[p[i] - 'a'] = max(cur, count[p[i] - 'a']);
}
return accumulate(count.begin(), count.end(), 0);
}
};
Java:
class Solution {
public int findSubstringInWraproundString(String p) {
int result = 0, n = p.length(), count[] = new int[26], cur = 1;
for (int i = 0; i < n; i++) {
if (i > 0 && (p.charAt(i) - p.charAt(i - 1) == 1 || p.charAt(i - 1) - p.charAt(i) == 25)) ++cur;
else cur = 1;
count[p.charAt(i) - 'a'] = Math.max(count[p.charAt(i) - 'a'], cur);
}
for (int c : count) result += c;
return result;
}
}
Python:
class Solution:
def findSubstringInWraproundString(self, p: str) -> int:
count, cur, result = [0] * 128, 1, 0
for i in range(len(p)):
cur = cur + 1 if i and (ord(p[i]) - ord(p[i - 1])) % 26 == 1 else 1
count[ord(p[i])] = max(count[ord(p[i])], cur)
return sum(count)