567. Permutation in String Given two stringss1ands2, returntrueifs2contains apermutationofs1, orfalseotherwise. In other words, returntrueif one ofs1's permutations is the substring ofs2. Example 1: Input:s1 = "ab", s2 = "eidbaooo"Output:trueExplanation:s2 contains one permutation of s1 ...
Given a strings, you can transform every letter individually to be lowercase or uppercase to create another string. Returna list of all possible strings we could create. Return the output inany order. Example 1: Input:s = "a1b2"Output:["a1b2","a1B2","A1b2","A1B2"] Example 2: ...
permutation 排列, palindrome 回文 判断一个string 能否组合成一个回文。思路参考Hint. 我用了HashMap, 计算如果character 的数目是odd, 最多只有一个时,为真。但从discussion中发现,有人的方法更巧妙,复杂度更低。贴过来参考。 还有一种方法,用HashSet. Java code: my code:(use map) publicbooleancanPermuteP...
public static void main(String[] args){ Solution sol = new Solution(); String s = "abc"; List<String> res = sol.permutation(s); System.out.println(res.size()); } }
If a palindromic permutation exists, we just need to generate the first half of the string. 本题难度系数很难,看了答案实现起来仍然出了不少错误。我开始想的是按照Permutations II的方法来做,找到后再进行筛选,选出满足palindrome 条件的数,实现起来比较麻烦,而且最后给出了内存不足,故打断了这种想法。