Practice when no one is watching...
1 書上說的第二種方法很贊,碰到很多連結串列移位的問題,都可以考慮這樣做。記得leetcode就有一道。
2 這道題目相對簡單,直接上程式碼
public static boolean StringShift1(String source,String test){ source = source + source; return source.contains(test); } public static boolean StringShift2(String source,String test){ char[] s1 = source.toCharArray(); for(int i=0;i<s1.length;i++){ char temp = s1[0]; for(int j=0;j<s1.length-1;j++){ s1[j]=s1[j+1]; } s1[s1.length-1]=temp; if(new String(s1).contains(test)){ return true; } } return false; }
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Practice when no one is watching...
思路 1 書上說的第二種方法很贊,碰到很多連結串列移位的問題,都可以考慮這樣做。記得leetcode就有一道。 2 這道題目相對簡單,直接上程式碼 public static boolean StringShift1(String source,String te
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