[Leetcode.467]Unique Substrings in Wraparound String
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问题描述
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.
解题思路:
方法:DP
1.charArray[i]里存的是以字符(char)i结尾的最长的连续子串的长度
如果新的子串长度大于旧的长度,说明有新的合法子串出现,这些新出现合法子串的数目是等于 新长度-旧长度
如:bcabc
扫描到第一个b时,charArray[‘b’]的值为0,新得到的已b结尾的最长连续子串长度是1(b),因此总数count加1并修改charArray[‘b’]的值为1(1-0),即加了’b‘这个子串
- 扫描到第一个c时,charArray[‘c’]的值为0,新得到的已c结尾的最长连续子串长度是2(bc),因此总数count再加2并修改charArray[‘c’]的值为2(2-0),即加了’c‘,’bc‘两个子串
- 扫描到a时,同上,count加1,修改了charArray[‘a’]的值为1,即加了’a‘这个子串
- 扫描到第二个b时,charArray[‘b’]的值为1,新得到的已b结尾的最长连续子串长度是2(ab),因此总数count再加1(2-1)并修改charArray[‘b’]的值为2,即加了’ab‘这个子串
- 扫描到第二个c时,charArray[‘c’]的值为2,新得到的已c结尾的最长连续子串长度是3(abc),因此总数count再加1(3-2)并修改charArray[‘b’]的值为3,即加了’abc‘这个子串
- 所以最后总数count是6,符合要求的子串是’b‘,’c‘,’bc‘,’a‘,’ab‘,’abc‘
代码(java):
public static int findSubstringInWraproundString(String p) { if(p == null || p.length() == 0){ return 0; } int[] charArray = new int[128]; int count = 0; for(int i = 0; i < p.length(); i++){ int cur = i - 1; while(cur >= 0 && (p.charAt(cur + 1) - p.charAt(cur) == 1 || p.charAt(cur + 1) - p.charAt(cur) == -25) ){ cur--; } if(i - cur > charArray[p.charAt(i)]){//i - cur表示以i结尾的最长连续子串的长度 count += i - cur - charArray[p.charAt(i)]; charArray[p.charAt(i)] = i - cur; } } return count; }
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