【leetcode】893. Groups of Special-Equivalent Strings

Algorithm

【leetcode】893. Groups of Special-Equivalent Strings

https://leetcode.com/problems/groups-of-special-equivalent-strings/

1）problem

You are given an array A of strings.

Two strings S and T are special-equivalent if after any number of moves, S == T.

A move consists of choosing two indices i and j with i % 2 == j % 2, and swapping S[i] with S[j].

Now, a group of special-equivalent strings from A is a non-empty subset S of A such that any string not in S is not special-equivalent with any string in S.

Return the number of groups of special-equivalent strings from A.

Example 1:

Input: ["a","b","c","a","c","c"]
Output: 3
Explanation: 3 groups ["a","a"], ["b"], ["c","c","c"]

Example 2:

Input: ["aa","bb","ab","ba"]
Output: 4
Explanation: 4 groups ["aa"], ["bb"], ["ab"], ["ba"]

Example 3:

Input: ["abc","acb","bac","bca","cab","cba"]
Output: 3
Explanation: 3 groups ["abc","cba"], ["acb","bca"], ["bac","cab"]

Example 4:

Input: ["abcd","cdab","adcb","cbad"]
Output: 1
Explanation: 1 group ["abcd","cdab","adcb","cbad"]

Note:

• 1 <= A.length <= 1000
• 1 <= A[i].length <= 20
• All A[i] have the same length.
• All A[i] consist of only lowercase letters.

2）answer

统计奇数位和偶数位的个字符出现的次数，如果 S0 串和 S1 串统计的结果相同，则它们是 special-equivalent 的。

3）solution

1. 将字符串拆分为两个子字符串，1个包含偶数索引字符，1个包含奇数字符串
2. 对两个子字符串进行排序（这样做是因为如果可以将字符串与另一个字符串交换，那么在排序时它们将彼此相等，因为它们必须具有相同的字符）
3. 将一对字符串插入集合中，这将跟踪唯一的“组”
4. 重新调整集合的大小

符合上面两个条件的是一组。

参考：https://blog.csdn.net/g_r_c/article/details/82079678

unordered_set使用方法

.find() 返回一个迭代器。这个迭代器指向和参数哈希值匹配的元素，如果没有匹配的元素，会返回这个容器的结束迭代器。
.end() 返回指向容器末尾位置的迭代器
.insert() 插入元素 

视频：

https://www.youtube.com/watch?v=WJ4NtyrakT0&feature=youtu.be

图片示例：

C++实现代码：

#include "pch.h"
#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
#include <algorithm>
using std::vector;
using std::string;
using std::unordered_set;

class Solution {
public:
    int numSpecialEquivGroups(vector<string>& A) {
        unordered_set<string> st;
        for (int i = 0; i < A.size(); ++i) {
            string odd = "";
            string even = "";
            for (int j = 0; j < A[i].size(); ++j) {
                // 奇偶位的值
                if (j % 2 == 0)
                    odd += A[i][j];
                else
                    even += A[i][j];
            }
            // 排序奇偶位的值
            sort(odd.begin(), odd.end());
            sort(even.begin(), even.end());
            st.insert(odd + even);

        }
        return st.size();
    }
};
int main()
{
    Solution solution;
    vector<string> A1 = { "abcd", "cdab", "adcb", "cbad" };
    vector<string> A2 = { "abc", "acb", "bac", "bca", "cab", "cba" };
    int res1 = solution.numSpecialEquivGroups(A1);
    int res2 = solution.numSpecialEquivGroups(A2);
}

Python实现：

def numSpecialEquivGroups(self,A):
    s = set()
    for w in A:
        even = ''.join(sorted(w[0::2]))
        odd  = ''.join(sorted(w[1::2]))
        s.add(odd+even)
    return len(s)