[IT技術30篇挑戰] 資料結構和演算法實作 DAY7 – 雜湊 (Hash) @地瓜大的飛翔旅程

章節連結

指令
LeetCode 練習記錄
系列文章

雜湊表是(Hash Table)由多個 key, value 對所組成的一個表。透過 Hash 雜湊值，就可以快速的找到對應的值，在操作上的時間複雜度為 1，無論是插入、刪除、更新……等。當資料跟排列順序無關，或是要計算出現次數(重複)值時，使用 Hash 是可用的方式。

指令

/*Hash*/
function HashTable(){
    let table = []
    let loseloseHashCode = (key) =>{
        let hash = 0
        for(let i=0;i<key.length;i++){
            hash += key.charCodeAt(i)
        }
        return hash % 37
    }

    this.put = (key,value)=>{
        let position = loseloseHashCode(key)
        console.log(position+'-'+key)
        table[position] = value
    }

    this.get = key =>{
        return table[loseloseHashCode(key)]
    }

    this.remove = key =>{
        table[loseloseHashCode(key)] = undefined
    }
}

let hash = new HashTable()
hash.put('Andy','andy@123.com')
hash.put('Wendy','wendy@123.com')
hash.put('Tom','tom@123.com')

console.log(hash.get('Andy'))
console.log(hash.get('Sandy'))

/*Hash*/

function HashTable(){

let table = []

let loseloseHashCode = (key) =>{

let hash = 0

for(let i=0;i<key.length;i++){

hash += key.charCodeAt(i)

}

return hash % 37

}

this.put = (key,value)=>{

let position = loseloseHashCode(key)

console.log(position+'-'+key)

table[position] = value

}

this.get = key =>{

return table[loseloseHashCode(key)]

}

this.remove = key =>{

table[loseloseHashCode(key)] = undefined

}

let hash = new HashTable()

hash.put('Andy','[email protected]')

hash.put('Wendy','[email protected]')

hash.put('Tom','[email protected]')

console.log(hash.get('Andy'))

console.log(hash.get('Sandy'))

LeetCode 練習記錄

// 242. Valid Anagram https://leetcode.com/problems/valid-anagram/
/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isAnagram = function(s, t) {
    if(s.length !== t.length) return false
    let hash = {}
    
    for(let i=0; i<s.length;i++){
        if(!hash[s[i]]){
            hash[s[i]] = 1
        }else{
            hash[s[i]] ++
        }
    }
    
    for(let i=0;i<t.length;i++){
        if(hash[t[i]]){
            hash[t[i]] --
            if(hash[t[i]].length<0){
                return false
            }
        }else{
            return false
        }
    }
    
    return true
};