标签：diff 另一个 long success public enc seq ++ forms
问题描述：
Given a string, we can "shift" each of its letter to its successive letter, for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence:
"abc" -> "bcd" -> ... -> "xyz"
Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence.
Example:
Input: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"],
Output:
[
["abc","bcd","xyz"],
["az","ba"],
["acef"],
["a","z"]
]
解题思路：
使用visited函数来判断该字符串是否已经被访问并且加入到某一个组内。
若加入则跳过
若未加入，则判断或寻找它的组员
需要注意的是：
在判断是否可以通过偏移来获得另一个字符串时，注意当被减数大于减数时如何判断：
s1[i] - ‘a‘ + (‘z‘ - s2[i]) +1;
代码：
class Solution { public: vector<vector<string>> groupStrings(vector<string>& strings) { vector<bool> visited(strings.size(), false); vector<vector<string>> ret; for(int i = 0; i < strings.size(); i++){ if(visited[i]) continue; visited[i] = true; vector<string> cur; cur.push_back(strings[i]); for(int j = i+1; j < strings.size(); j++){ if(visited[j]) continue; if(canShift(strings[i], strings[j])){ cur.push_back(strings[j]); visited[j] = true; } } ret.push_back(cur); } return ret; } private: bool canShift(string s1, string s2){ if(s1.size() != s2.size()){ return false; } if(s1.size() == 1) return true; int diff = s1[0] >= s2[0] ? s1[0] - s2[0] : s1[0] - ‘a‘ + (‘z‘ - s2[0]) + 1; for(int i = 1; i < s1.size(); i++){ int curdiff = s1[i] >= s2[i] ? s1[i] - s2[i] : s1[i] - ‘a‘ + (‘z‘ - s2[i]) +1; if(curdiff != diff) return false; } return true; } };
标签：diff 另一个 long success public enc seq ++ forms
原文地址：https://www.cnblogs.com/yaoyudadudu/p/9241471.html
