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  • 839. Similar String Groups

    package LeetCode_839
    
    import java.util.*
    import kotlin.collections.HashSet
    
    /**
     * 839. Similar String Groups
     * https://leetcode.com/problems/similar-string-groups/
     * Two strings X and Y are similar if we can swap two letters (in different positions) of X, so that it equals Y.
     * Also two strings X and Y are similar if they are equal.
    For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar,
    but "star" is not similar to "tars", "rats", or "arts".
    Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}.
    Notice that "tars" and "arts" are in the same group even though they are not similar.
    Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.
    We are given a list strs of strings where every string in strs is an anagram of every other string in strs.
    How many groups are there?
    
    Example 1:
    Input: strs = ["tars","rats","arts","star"]
    Output: 2
    
    Example 2:
    Input: strs = ["omv","ovm"]
    Output: 1
    
    Constraints:
    1. 1 <= strs.length <= 100
    2. 1 <= strs[i].length <= 1000
    3. sum(strs[i].length) <= 2 * 10^4
    4. strs[i] consists of lowercase letters only.
    5. All words in strs have the same length and are anagrams of each other.
     * */
    class Solution {
        /*
        * solution: BFS, do bfs for each string to check current and next if similar,
        * Time:O(n^2*l), l is length of str, Space:O(n),
        * */
        fun numSimilarGroups(strs: Array<String>): Int {
            if (strs.size < 2) {
                return strs.size
            }
            var group = 0
            val visited = HashSet<String>()
            val queue = LinkedList<String>()
            for (str in strs) {
                if (visited.contains(str)) {
                    continue
                }
                visited.add(str)
                queue.offer(str)
                group++
                while (queue.isNotEmpty()) {
                    val cur = queue.poll()
                    //cur string compare with next string
                    for (item in strs) {
                        if (visited.contains(item)) {
                            continue
                        }
                        var diff = 0
                        //if the number of different char equals to 2, is similar;
                        //all word have the same length
                        for (i in cur.indices) {
                            if (cur[i] != item[i]) {
                                diff++
                            }
                        }
                        //put similar string into queue to check next level
                        if (diff == 2) {
                            visited.add(item)
                            queue.offer(item)
                        }
                    }
                }
            }
    
            return group
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/14280042.html
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