LeetCode 54: Spiral Matris – JavaScript \u00c7\u00f6z\u00fcm\u00fc<\/p>\n
Merhaba! Fatih Soysal olarak, bug\u00fcn sizlerle LeetCode’un olduk\u00e7a pop\u00fcler bir probleminden, 54. problem olan Spiral Matris probleminden ve JavaScript \u00e7\u00f6z\u00fcm\u00fcnden bahsedece\u011fim. Bu problemde, verilen bir matrisi spiral \u015fekilde dola\u015farak elemanlar\u0131n\u0131 s\u0131ral\u0131 bir dizi olarak d\u00f6nd\u00fcrmemiz gerekiyor. \u00d6ncelikle problemi daha iyi anlamak i\u00e7in bir \u00f6rne\u011fe bakal\u0131m. <\/p>\n
\u00d6rnek:<\/strong><\/p>\n Girdi: \u00c7\u0131kt\u0131: Bu \u00f6rnekte, matris spiral \u015fekilde dola\u015f\u0131l\u0131yor ve elemanlar s\u0131rayla diziye ekleniyor. Peki bu problemi nas\u0131l \u00e7\u00f6zebiliriz?<\/p>\n Bu problemi \u00e7\u00f6zmek i\u00e7in birka\u00e7 farkl\u0131 yakla\u015f\u0131m kullan\u0131labilir. Ancak en yayg\u0131n ve anla\u015f\u0131l\u0131r y\u00f6ntemlerden biri, matrisin s\u0131n\u0131rlar\u0131n\u0131 takip ederek spiral bir \u015fekilde dola\u015fmakt\u0131r. Bu y\u00f6ntemi JavaScript ile kodlayal\u0131m. \u00d6ncelikle, matrisin boyutlar\u0131n\u0131 ve s\u0131n\u0131rlar\u0131n\u0131 tutan de\u011fi\u015fkenler tan\u0131ml\u0131yoruz. Daha sonra, d\u00f6ng\u00fcler kullanarak matrisin d\u0131\u015f kenar\u0131n\u0131 takip ediyor ve elemanlar\u0131 diziye ekliyoruz. Her bir turda, s\u0131n\u0131rlar\u0131 daraltarak matrisin i\u00e7ine do\u011fru ilerliyoruz. Bu i\u015flem, matrisin ortas\u0131na ula\u015fana kadar tekrarlan\u0131yor.<\/p>\n Bu kod, matrisi d\u00f6rt y\u00f6nde (sa\u011f, a\u015fa\u011f\u0131, sol, yukar\u0131) dola\u015farak elemanlar\u0131 s\u0131rayla diziye ekler. Sonu\u00e7 olarak, spiral \u015fekilde dola\u015f\u0131lm\u0131\u015f matrisin elemanlar\u0131n\u0131 i\u00e7eren bir dizi elde ederiz. Bu kodun anla\u015f\u0131l\u0131rl\u0131\u011f\u0131 i\u00e7in, de\u011fi\u015fkenlerin a\u00e7\u0131klay\u0131c\u0131 isimler kullan\u0131lm\u0131\u015f ve kodun her b\u00f6l\u00fcm\u00fc a\u00e7\u0131k bir \u015fekilde yorumlanm\u0131\u015ft\u0131r. Ayr\u0131ca, d\u00f6ng\u00fcler ve ko\u015fullar\u0131n mant\u0131\u011f\u0131n\u0131 takip etmek olduk\u00e7a kolayd\u0131r. <\/p>\n Bu \u00e7\u00f6z\u00fcm\u00fcn zaman karma\u015f\u0131kl\u0131\u011f\u0131 O(m*n) , burada m ve n s\u0131ras\u0131yla matrisin sat\u0131r ve s\u00fctun say\u0131s\u0131d\u0131r. Bu, matrisin her eleman\u0131n\u0131n bir kez ziyaret edilmesi gerekti\u011fi anlam\u0131na gelir. Uzun ve karma\u015f\u0131k matrislerde bile bu \u00e7\u00f6z\u00fcm olduk\u00e7a verimlidir. <\/p>\n Umar\u0131m bu makale size yard\u0131mc\u0131 olmu\u015ftur. Daha fazla LeetCode \u00e7\u00f6z\u00fcm\u00fc ve di\u011fer yaz\u0131l\u0131m geli\u015ftirme konular\u0131 i\u00e7in fatihsoysal.com<\/a> adresini ziyaret edebilirsiniz. Ayr\u0131ca, JavaScript \u00f6\u011frenmek i\u00e7in \u00e7e\u015fitli online kaynaklardan faydalanabilirsiniz. \u00d6rne\u011fin, Mozilla Developer Network<\/a> harika bir ba\u015flang\u0131\u00e7 noktas\u0131 olabilir. <\/p>\n #LeetCode #JavaScript #SpiralMatrix #Algoritma #Programlama #Kodlama #FatihSoysal #Problem\u00c7\u00f6z\u00fcm\u00fc #Yaz\u0131l\u0131mGeli\u015ftirme<\/p>\n","protected":false},"excerpt":{"rendered":"LeetCode 54: Spiral Matris – JavaScript \u00c7\u00f6z\u00fcm\u00fc Merhaba! Fatih Soysal olarak, bug\u00fcn sizlerle LeetCode’un olduk\u00e7a pop\u00fcler bir probleminden,…","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"csco_page_header_type":"","csco_page_load_nextpost":"","csco_page_subscribe_form":"","csco_page_contact_form":"","footnotes":""},"categories":[1],"tags":[],"class_list":{"0":"post-9101","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"\n[[1,2,3],[4,5,6],[7,8,9]]<\/code><\/p>\n[1,2,3,6,9,8,7,4,5]<\/code><\/p>\nProblem \u00c7\u00f6z\u00fcm\u00fc:<\/h2>\n
\n\/**\n * @param {number[][]} matrix\n * @return {number[]}\n *\/\nvar spiralOrder = function(matrix) {\n const result = [];\n let top = 0, bottom = matrix.length - 1;\n let left = 0, right = matrix[0].length - 1;\n let dir = 0; \/\/ 0: right, 1: down, 2: left, 3: up\n\n while (top <= bottom && left <= right) {\n if (dir === 0) {\n for (let i = left; i <= right; i++) {\n result.push(matrix[top][i]);\n }\n top++;\n } else if (dir === 1) {\n for (let i = top; i <= bottom; i++) {\n result.push(matrix[i][right]);\n }\n right--;\n } else if (dir === 2) {\n for (let i = right; i >= left; i--) {\n result.push(matrix[bottom][i]);\n }\n bottom--;\n } else if (dir === 3) {\n for (let i = bottom; i >= top; i--) {\n result.push(matrix[i][left]);\n }\n left++;\n }\n dir = (dir + 1) % 4;\n }\n return result;\n};\n<\/code><\/pre>\n#Etiketler<\/h2>\n