LeetCode 467. Unique Substrings in Wraparound String

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题目要求

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 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.

题意解析

计算一个字符串内,有多少字串也是”…zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd….”的子串。

解法分析

这道题最直接的方法就是遍历,时间复杂度是O(n2),这个是没有通过的。

看了Discuss之后毛瑟顿开。针对这道题来说,以一个字符作为字符串的结尾,只要知道连续的长度有多少,那么以这个字符结尾的子串就有多少个,这样只需要遍历一遍就可以了。

解题代码

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public int findSubstringInWraproundString(String p) {
// count[i] is the maximum unique substring end with ith letter.
// 0 - 'a', 1 - 'b', ..., 25 - 'z'.
int[] count = new int[26];
// store longest contiguous substring ends at current position.
int maxLengthCur = 0;
for (int i = 0; i < p.length(); i++) {
if (i > 0
&& (p.charAt(i) - p.charAt(i - 1) == 1 || (p.charAt(i - 1)
- p.charAt(i) == 25))) {
maxLengthCur++;
} else {
maxLengthCur = 1;
}
int index = p.charAt(i) - 'a';
count[index] = Math.max(count[index], maxLengthCur);
}
// Sum to get result
int sum = 0;
for (int i = 0; i < 26; i++) {
sum += count[i];
}
return sum;
}