摘要：這是我在平時(shí)有時(shí)間的時(shí)候做的一些算法上的題目想看更新請(qǐng)移步這里題目描述解法這個(gè)問(wèn)題當(dāng)時(shí)拿到的時(shí)候是完全沒(méi)有思路的，后面上網(wǎng)查詢(xún)了一下這個(gè)題目，知道了使用斐波那契數(shù)列就能夠解這道題目，，，當(dāng)然百度作業(yè)幫上面也有相應(yīng)的解法，套路就是題目為一

想看更新請(qǐng)移步這里

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.

這個(gè)問(wèn)題當(dāng)時(shí)拿到的時(shí)候是完全沒(méi)有思路的，后面上網(wǎng)查詢(xún)了一下這個(gè)題目，知道了使用
斐波那契數(shù)列就能夠解這道題目，`F(0)=1，F(xiàn)(1)=1, F(n)=F(n-1)+F(n-2)（n>=2
，n∈N*）`當(dāng)然百度作業(yè)幫上面也有相應(yīng)的解法，套路就是

題目為一次可以走一步或兩步:
a[1]=1;a[2]=2;
假設(shè)i的范圍為3~n（n 為樓梯數(shù)），關(guān)系式為：
a[i]=a[i-1]+a[i-2];

若題目為一次可以走一步或兩步或三步：
a[1]=1;a[2]=2;a[3]=4;
假設(shè)i的范圍為4~n（n為樓梯數(shù)），關(guān)系式為：
a[i]=a[i-1]+a[i-2]+a[i-3]; 

但是當(dāng)初看到這個(gè)數(shù)列的公式，第一想法是使用遞歸，然后，由于堆棧溢出，失敗了
再看看后面的百度作業(yè)幫的那個(gè)公式，再聯(lián)想一下數(shù)列，于是想到了使用數(shù)列這樣的
結(jié)構(gòu)來(lái)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

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
這是我的解法，寫(xiě)的真的很丑，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]];
}
然后是大神的解法，使用的是位運(yùn)算（說(shuō)實(shí)話(huà)，還不是很懂位運(yùn)算，如果有什么見(jiàn)解，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(jiǎn)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(gè)代碼字?jǐn)?shù)限制，只能在30個(gè)字符以?xún)?nèi)，使用了ES6的箭頭函數(shù)
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(duì)簡(jiǎn)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
我的解法
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(diǎn)意思，細(xì)心的人就會(huì)發(fā)現(xiàn)給出的例子里面可以看出一些等差數(shù)列的影子，那么我們
就可以按照等差數(shù)列的方法來(lái)解這道題了，等差數(shù)列求和公式：（首項(xiàng) + 末項(xiàng)）* 項(xiàng)數(shù) / 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：這里面我們需要求項(xiàng)數(shù)N和末項(xiàng)An，至于N怎么求的，那就要靠自己去發(fā)現(xiàn)規(guī)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
我的解法
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);
};
在這里看了下大神的解法，讓我對(duì)reduce函數(shù)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
我的解法
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
我的解法
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(shù)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(jié)構(gòu)，用于存儲(chǔ)數(shù)字類(lèi)似于：
// 807這個(gè)數(shù)字的存儲(chǔ)結(jié)構(gòu)
var num8 = {
  val: 8,
  next: null,
}

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

var num7 = {
  val: 7,
  next: num0,
}
// 807這個(gè)數(shù)字就能夠用這樣的結(jié)構(gòu)得到
就只是用一個(gè)新的linked list來(lái)儲(chǔ)存相加後的結(jié)果
要注意的就是list1跟list2長(zhǎng)度可能不一樣
另外就是相加後可能比9還大，需要考慮進(jìn)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(chǔ)存輸出的結(jié)果，因?yàn)閘ist的指針要不斷往後移，因此用一個(gè)假節(jié)點(diǎn)方便操作
    var result = list; // 使用一個(gè)ListNode來(lái)儲(chǔ)存相加的結(jié)果

    var sum,carry = 0; // carry用來(lái)處理進(jìn)位

    //當(dāng) list1, list2 都沒(méi)有值，而且carry也為0的時(shí)候才結(jié)束迴圈
    while(l1 || l2 || carry > 0){
        sum = 0;

        // list1與list2長(zhǎng)度可能不同，分開(kāi)處理
        if(l1!== null){
            sum += l1.val;
            l1 = l1.next;
        }

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

        // 如果之前有進(jìn)位，carry = 1；沒(méi)有的話(huà)carry = 0
        sum += carry;
        list.next = new ListNode(sum%10); //相加如果超過(guò)9，只能留下個(gè)位數(shù)放入結(jié)果list，十位數(shù)的地方進(jìn)位
        carry = parseInt(sum/10);

        // list指標(biāo)向後
        list = list.next;
    }
    // 因?yàn)榈谝粋€(gè)節(jié)點(diǎn)為假節(jié)點(diǎn)，跳過(guò)
    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
我的解法
"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(jìn)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
我的解法
要比較兩個(gè)字串裡面的字元是否相同，首先可以判斷長(zhǎng)度是否相等，不相等就可以直接判定
為false 接下來(lái)將重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;
};