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  • LeetCode Repeated String Match

    原题链接在这里:https://leetcode.com/problems/repeated-string-match/description/

    题目:

    Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1.

    For example, with A = "abcd" and B = "cdabcdab".

    Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").

    Note:
    The length of A and B will be between 1 and 10000.

    题解:

    A重复append自己知道比B长,看B是否包含在内,若包含返回当前count.

    若没有再append次A. 若B能是substring, 这个长度已经可以包含所有可能性. B的开头在0到A.length()-1的任何位置都足够盖住B.

    Time Complexity: O(A.length()+B.length()). create个长度为A.length()+B.length()的string. 再用B找index.

    Space: O(A.length()+B.length()).

    AC Java:

    class Solution {
        public int repeatedStringMatch(String A, String B) {
            int count = 0;
            StringBuilder sb = new StringBuilder();
            while(sb.length() < B.length()){
                sb.append(A);
                count++;
            }
            
            if(sb.indexOf(B) >= 0){
                return count;
            }else if(sb.append(A).indexOf(B) >= 0){
                return count+1;
            }else{
                return -1;
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/7684884.html
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