一些作着玩的題

這是我在平時有時間的時候作的一些算法上的題目

想看更新請移步這裏

題目： Climbing Stairs

描述

You are climbing a stair case.It takes n steps to reach to the top

Each time you can either climb 1 or 2 steps. In how many distinct 
ways can you climb to the top?

Note: Given n will be a positive integer.

解法

這個問題當時拿到的時候是徹底沒有思路的，後面上網查詢了一下這個題目，知道了使用
斐波那契數列就可以解這道題目，`F(0)=1，F(1)=1, F(n)=F(n-1)+F(n-2)（n>=2
，n∈N*）`固然百度做業幫上面也有相應的解法，套路就是

題目爲一次能夠走一步或兩步:
a[1]=1;a[2]=2;
假設i的範圍爲3~n（n 爲樓梯數），關係式爲：
a[i]=a[i-1]+a[i-2];

若題目爲一次能夠走一步或兩步或三步：
a[1]=1;a[2]=2;a[3]=4;
假設i的範圍爲4~n（n爲樓梯數），關係式爲：
a[i]=a[i-1]+a[i-2]+a[i-3];

可是當初看到這個數列的公式，第一想法是使用遞歸，而後，因爲堆棧溢出，失敗了
再看看後面的百度做業幫的那個公式，再聯想一下數列，因而想到了使用數列這樣的
結構來解決

/**
 * @param {number} n
 * @return {number}
 */
var climbStairs = function(n) {
    if (isNaN(n)) return -1;
    if (n<=0) return -1;
    if (n<=2) return n;
    if (n>=3) {
        var res = [1,2];
        for(var i=2;i<n;i++){  
            res[i] = res[i-1] + res[i-2];  
        }  
        return res[n-1];
    }
};

---

題目： Find the odd int

描述

Given an array, find the int that appears an odd number of times.

There will always be only one integer that appears an odd number
 of times.

解法

這是個人解法，寫的真的很醜，勿噴哦

function findOdd(A) {
  //happy coding!
  let indecies = [];
  
  A.forEach(a => {
    let idx = A.indexOf(a);
    let index = [];
    while (idx !== -1) {
      index.push(idx);
      idx = A.indexOf(a, idx + 1);
    }
    indecies.push(index);
  });
  const a = indecies.filter(index => (index.length % 2 !== 0));
  return A[a[0][0]];
}

而後是大神的解法，使用的是位運算（說實話，還不是很懂位運算，若是有什麼看法，歡迎留言）

function findOdd(A) {
  //happy coding!
  return A.reduce((a, b) => a^b);
}

---

題目：You're a square!

描述

Given an integral number, determine if it's a square number:
isSquare(-1) => false
isSquare( 3) => false
isSquare( 4) => true
isSquare(25) => true
isSquare(26) => false

解法

這道就簡單了

var isSquare = function(n){
  return Number.isInteger(Math.sqrt(n));
}

---

題目：Strings to numbers

描述

You are given an array of numbers in string form. Your task is to 
convert this to an array of numbers.

Your function can only be a maximum of 30 characters long 
(not including whitespaces)! I have limited the char count 
because there is a very short and easy way to achieve this task.

Here is an example of what your function needs to return:

['1','2','3'] => [1,2,3]
Edge Cases:

1 - If your function comes up against a value that isn't a number
 its place in the array must be substituted with NaN.

2 - An empty array must return an empty array.

解法

這道題主要是有一個代碼字數限制，只能在30個字符之內，使用了ES6的箭頭函數

var convert = a => a.map(n => n*1)

---

題目：Thinkful - List and Loop Drills: Inverse Slicer

描述

You're familiar with list slicing in Python and know, 
for example, that:
>>> ages = [12, 14, 63, 72, 55, 24]
>>> ages[2:4]
[63, 72]
>>> ages[2:]
[63, 72, 55, 24]
>>> ages[:3]
[12, 14, 63]

write a function inverse_slice() that takes three arguments: 
a list items, an integer a and an integer b. The function should
 return a new list with the slice specified by items[a:b] excluded
For example:
>>>inverse_slice([12, 14, 63, 72, 55, 24], 2, 4)
[12, 14, 55, 24]

The input will always be a valid list, a and b will always be 
different integers equal to or greater than zero, but they may 
be zero or be larger than the length of the list.

解法

這道題也相對簡單

function inverseSlice(items, a, b) {
  return items.filter((item, index) => !((index >= a) && (index < b)));
}

---

題目：Get all array elements except those with specified indexes

描述

Extend the array object with a function to return all elements
 of that array, except the ones with the indexes passed in the
  parameter.

For example:

var array = ['a', 'b', 'c', 'd', 'e'];
var array2 = array.except([1,3]);
// array2 should contain ['a', 'c', 'e'];
The function should accept both array as parameter but also a
 single integer, like this:

var array = ['a', 'b', 'c', 'd', 'e'];
var array2 = array.except(1);
// array2 should contain ['a', 'c', 'd', 'e'];

解法

個人解法

Array.prototype.except = function(keys)
{
  if (typeof keys === 'number') {
    return this.filter((item, index) => index !== keys);
  }
  if (Array.isArray(keys)) {
    return this.filter((item, index) => !keys.includes(index));
  }
  return -1;
}

大神的解法

Array.prototype.except = function(keys)
{
  if(!Array.isArray(keys)) keys = [keys];
  return this.filter((a,i) => !keys.includes(i));
}

---

題目：Sum of a Sequence [Hard-Core Version]

描述

The task is simple to explain: simply sum all the numbers from 
the first parameter being the beginning to the second parameter
 being the upper limit (possibly included), going in steps 
 expressed by the third parameter:

sequenceSum(2,2,2) === 2
sequenceSum(2,6,2) === 12 // 2 + 4 + 6
sequenceSum(1,5,1) === 15 // 1 + 2 + 3 + 4 + 5
sequenceSum(1,5,3) === 5 // 1 + 4
If it is an impossible sequence (with the beginning being larger
 the end and a positive step or the other way around), just 
 return 0. See the provided test cases for further examples :)

Note: differing from the other base kata, much larger ranges are
going to be tested, so you should hope to get your algo optimized
and to avoid brute-forcing your way through the solution.

解法

這道題有點意思，細心的人就會發現給出的例子裏面能夠看出一些等差數列的影子，那麼咱們
就能夠按照等差數列的方法來解這道題了，等差數列求和公式：（首項 + 末項）* 項數 / 2

function sequenceSum(begin, end, step){
  //your code here 
  const n = Math.floor((end - begin) / step) + 1;
  if (n<0) return 0; 
  const An = begin + ((n-1) * step);
  return n * (begin + An) / 2;
}

注意：這裏面咱們須要求項數N和末項An，至於N怎麼求的，那就要靠本身去發現規律

---

題目：Array.prototype.size()

描述

Implement Array.prototype.size() - without .length !

Implement Array.prototype.size(), which should simply return
the length of the array. But do it entirely without using
Array.prototype.length!
Where .length is a property, .size() is a method.

Rules

Because it is quite impossible to disable [].length, and because
filtering for "length" is an iffy proposition at best,
THIS KATA WORKS ON THE HONOUR SYSTEM.
You may cheat. But you may have trouble sleeping. Or $DEITY may
kill a puppy.

You need not support sparse arrays (but you may!). All testing
will be done with dense arrays.
Values will not be undefined. You need only support actual, real
arrays.

Your method needs to be read only. Arguments must be ignored. The
this object must not be modified.

解法

個人解法

Array.prototype.size = function () {
  var i = 0;
  while (this[i] !== undefined) i++;
  return i;
};

大神的解法

Array.prototype.size = function() {
  return this.reduce(r => r + 1, 0);
};

在這裏看了下大神的解法，讓我對reduce函數又有了新的理解

---

題目：Find the smallest integer in the array

描述

Find the smallest integer in the array.

Given an array of integers your solution should find the smallest
integer. For example:
Given [34, 15, 88, 2] your solution will return 2
Given [34, -345, -1, 100] your solution will return -345

You can assume, for the purpose of this kata, that the supplied
array will not be empty.

解法

個人解法

function findSmallestInt(args) {
  return args.sort((a, b) => {
    return a - b;
  })[0];
}

大神的解法

function findSmallestInt(args) {
  return Math.min(...args);
}

---

題目：Replace With Alphabet Position

描述

Welcome. In this kata you are required to, given a string,
replace every letter with its position in the alphabet. If 
anything in the text isn't a letter, ignore it and don't 
return it. a being 1, b being 2, etc. As an example:

alphabet_position("The sunset sets at twelve o' clock.")
Should return "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 
5 12 22 5 15 3 12 15 3 11" (As a string.)

解法

個人解法

function alphabetPosition(text) {
  let alphabet = {};
  let arr = [];
  const lowCaseText = text.toLowerCase();
  for (let i = 0; i < 26; i++) {
    alphabet[(String.fromCharCode((65+i))).toLowerCase()] = (i + 1);
  }
  for (let j = 0; j < text.length; j++) {
    if (alphabet[lowCaseText[j]]) {
      arr.push(alphabet[lowCaseText[j]]);
    }
  }
  return arr.join(' ');
}

大神的解法

function alphabetPosition(text) {
  var result = "";
  for (var i = 0; i < text.length; i++){
    var code = text.toUpperCase().charCodeAt(i)
    if (code > 64 && code < 91) result += (code - 64) + " ";
  }

  return result.slice(0, result.length-1);
}

---

題目：GetSum

描述

Given two integers, which can be positive and negative, find the
 sum of all the numbers between including them too and return it. 
 If both numbers are equal return a or b.

Note! a and b are not ordered!

Example: 
GetSum(1, 0) == 1   // 1 + 0 = 1
GetSum(1, 2) == 3   // 1 + 2 = 3
GetSum(0, 1) == 1   // 0 + 1 = 1
GetSum(1, 1) == 1   // 1 Since both are same
GetSum(-1, 0) == -1 // -1 + 0 = -1
GetSum(-1, 2) == 2  // -1 + 0 + 1 + 2 = 2

解法

依然是等差數列

function GetSum( a,b ) {
  //Good luck!
  return (a + b) * (Math.abs(b - a) + 1) / 2;
}

---

題目：Add Two Numbers

描述

You are given two linked lists representing two non-negative 
numbers. The digits are stored in reverse order
and each of their nodes contain a single digit. Add the two 
numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8

解法

javascript linked list structure

linked list 是一種鏈表結構，用於存儲數字相似於：

// 807這個數字的存儲結構
var num8 = {
  val: 8,
  next: null,
}

var num0 = {
  val: 0,
  next: num8,
}

var num7 = {
  val: 7,
  next: num0,
}
// 807這個數字就可以用這樣的結構獲得

就只是用一個新的linked list來儲存相加後的結果
要注意的就是list1跟list2長度可能不一樣
另外就是相加後可能比9還大，須要考慮進位的情況

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    var list = new ListNode(0); //儲存輸出的結果，因為list的指針要不斷往後移，所以用一個假節點方便操做
    var result = list; // 使用一個ListNode來儲存相加的結果

    var sum,carry = 0; // carry用來處理進位

    //當 list1, list2 都沒有值，並且carry也為0的時候才結束迴圈
    while(l1 || l2 || carry > 0){
        sum = 0;

        // list1與list2長度可能不一樣，分開處理
        if(l1!== null){
            sum += l1.val;
            l1 = l1.next;
        }

        if(l2!==null){
            sum += l2.val;
            l2 = l2.next;
        }

        // 若是以前有進位，carry = 1；沒有的話carry = 0
        sum += carry;
        list.next = new ListNode(sum%10); //相加若是超過9，只能留下個位數放入結果list，十位數的地方進位
        carry = parseInt(sum/10);

        // list指標向後
        list = list.next;
    }
    // 因為第一個節點為假節點，跳過
    return result.next;
}

// The complete code for the List constructor is:
function List() {
 List.makeNode = function() { 
  return {data: null, next: null}; 
 }; 
 
 this.start = null; 
 this.end = null; 
 
 this.add = function(data) { 
  if (this.start === null) { 
   this.start = List.makeNode(); 
   this.end = this.start; 
  } else { t
   this.end.next = List.makeNode(); 
   this.end = this.end.next; 
  } ; 
  this.end.data = data; 
 }; 

 this.delete = function(data) { 
  var current = this.start; 
  var previous = this.start; 
  while (current !== null) { 
   if (data === current.data) { 
    if (current === this.start) { 
     this.start = current.next; 
     return; 
    } 
    if (current === this.end) 
                      this.end = previous;
    previous.next = current.next; return; 
    }
    previous = current; 
    current = current.next; 
   }
 }; 

 this.insertAsFirst = function(d) { 
  var temp = List.makeNode(); 
  temp.next = this.start; 
  this.start = temp; 
  temp.data = d; 
 }; 

 this.insertAfter = function(t, d) { 
  var current = this.start; 
  while (current !== null) { 
   if (current.data === t) { 
    var temp = List.makeNode();
    temp.data = d; 
    temp.next = current.next; 
    if (current === this.end) this.end = temp;
    current.next = temp; 
    return; 
   } 
   current = current.next; 
   }
  };

  this.item = function(i) { 
   var current = this.start; 
   while (current !== null) { 
    i--; 
    if (i === 0) return current; 
    current = current.next; 
   } 
   return null; 
  }; 

 this.each = function(f) {
  var current = this.start;
  while (current !== null) { 
   f(current); 
   current = current.next; 
  } 
 };
}

---

題目：Reverse String

描述

Write a function that takes a string as input and returns the string reversed.
Example: Given s = "hello", return "olleh".

解法

個人解法

"hello" -> ['h','e','l','l','o'] -> 'o'+'l'+'l'+'e'+'h'

var reverseString = function(s) {
    var result = "";
    var ary = s.split("");

    for(var i = ary.length-1 ; i >= 0 ; i--){
        result = result + ary[i];
    }
    return result;
};

進階解法

未交換前 ['h','e','l','l','o']
第1次交換 ['h','e','l','l','o'] o,h互換 ['o','e','l','l','h']

第2次交換 ['o','e','l','l','h'] e,l互換 ['o','l','l','e','h']

var reverseString = function(s) {
    var result = "";
    var ary = s.split("");
    for(var i = 0, max = (ary.length-1)/2 ; i < max   ; i++){
        var temp = ary[i];
        ary[i] = ary[ary.length - 1 - i];
        ary[ary.length - 1 - i] = temp;
    }
    return ary.join("");
};

---

題目：Valid Anagram

描述

Given two strings s and t, write a function to determine if t is 
an anagram of s.

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.

Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you 
adapt your solution to such case?

解法

個人解法

要比較兩個字串裡面的字元是否相同，首先能夠判斷長度是否相等，不相等就能夠直接斷定

為false 接下來將從新排序後的字串比較是否相等。

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isAnagram = function(s, t) {
    if(s.length !== t.length) return false;

    var s1 = s.split("").sort().join("");
    var t1 = t.split("").sort().join("");

    return s1 === t1;
};