everyday-algorithms

每日手写及算法

算法目标：10分钟之内运行完成
手写目标：5分钟运行完成

doc：整理文档 solv：解题

提示：适应一下没有代码提示的编辑器！！！

Day1

1.手写算法 https://leetcode-cn.com/problems/add-to-array-form-of-integer/

//  时间复杂度： O(max(n,logk)) 
funciton resolve(num, k) {
  const len = num.length;
  let i = len - 1;
  const result = []
  while(i >= 0) {
    const sum = num[i] + k % 10;
    result.push(sum % 10);
    k = Math.floor(k / 10);
    if(sum >= 10) {
      k++;
    }
    i--;
  }
  while(k) {
    result.push(k % 10);
    k = Math.floor(k / 10);
  }
  return result.reverse();
}

2.编程题

//1.手写函数柯里化
function curry(func) {
  //此处补全
}
function sum(a, b, c) {
  return a + b + c;
}

let curriedSum = curry(sum);

alert(curriedSum(1, 2, 3)); // 6, still callable normally
alert(curriedSum(1)(2, 3)); // 6, currying of 1st arg
alert(curriedSum(1)(2)(3)); // 6, full currying

Day2

1.手写算法 https://leetcode-cn.com/problems/shortest-distance-to-a-character

2.编程题实现symbol polyfill //题解：如果浏览器不支持情况下写出让代码让浏览器支持symbol

Day3

3月30日：做题时间未达标，第一题15m，由于没有代码提示，导致没有发现API错误；第二题，10m中内做题失败，代码冗余，代码水平太差

1.编程题：https://bigfrontend.dev/zh/problem/implement-Promise-allSettled

/**
 * @param {Array<any>} promises - notice that input might contains non-promises
 * @return {Promise<Array<{status: 'fulfilled', value: any} | {status: 'rejected', reason: any}>>}
 */
function allSettled(promises) {
  if(!Array.isArray(promises)) {
    throw Error("invalid array");
  }
  if(promises.length == 0) {
    return Promise.resolve([])
  }
  let count = 0;
  const len = promises.length;
  // your code here
  const result = new Array(len);
  return new Promise(resolve => {
    promises.forEach((promise, i) => {
      Promise.resolve(promise).then(
        res => {
          result[i] = {
            status: "fulfilled",
            value: res
          };
        } ,
        error=> {
          result[i] = {
            status: "rejected",
            reason: error
          };
         
        }
      ).finally(() => {
          count++;
          if(count === len) {
            resolve(result);
          }
      })
    })
  })
  
}

2.算法题：https://leetcode-cn.com/problems/design-a-stack-with-increment-operation/

/**
 * @param {number[]} num
 * @param {number} k
 * @return {number[]}
 */
var addToArrayForm = function(num, k) {
    const m = num.length;
    let i = m -1;
    const result = [];
    while(i>=0) {
        const sum = num[i] +  k % 10;
        result.push(sum % 10);
        k = Math.floor(k / 10);
        if(sum >= 10) k++
        i--;
    }
    while(k) {
        result.push(k % 10);
        k = Math.floor(k / 10)
    }
    return result.reverse()
    
};

Day4

3月31日：
- 手写题：（12分钟）手写关于a、b都为0时，使用===无法区分，借助js中除数为0时不会抛出异常，对结果进行比较，-Infinity !== Infinity 不同（Object.is特别注意0及NaN的判断）
- 算法题：毁灭吧，思路都知道，就是写不出来。
  - 测试通过，但是代码逻辑不清晰，解题过程中，部分逻辑混乱，比如数字的获取与push字符及'['的先后选项 -- 代码水平不高
  - 需要再复习

1.手写题：https://bigfrontend.dev/zh/problem/implement-Object.is

function is(a, b) {
  // your code here
  if(Number.isNaN(a) && Number.isNaN(b)) {
    return true;
  }
  if(a === 0 && b === 0) {
    return 1 / a === 1 / b;
  }
  return a === b;
}

2.算法题：https://leetcode-cn.com/problems/decode-string/

/**
 * @param {string} s
 * @return {string}
 */
var decodeString = function(s) {
    const len = s.length;
    let i = 0;
    const stack = [];
    let num = "";
    while(i < len) {
        const c = s[i];
        if(c === ']') {
            let char = stack.pop();
            let str = "";
            while(char !== '[') {
                str = char + str;
                char = stack.pop();
            }
            const n = stack.pop();
            str = str.repeat(n);
            stack.push(str);
        } else if(isNumber(c)) {
            num += c;
        } else {
             if(num) {
                stack.push(Number(num));
                num = "";
            }
            stack.push(c);
        }
        i++;
    }
    return stack.join("");
};

function isNumber(c) {
    return !Number.isNaN(Number(c));
}

Day5

1.算法题：https://leetcode-cn.com/problems/implement-queue-using-stacks/

var MyQueue = function() {
    // 队首出队stack
    this._de = [];
    // 入队stack
    this._en = []
};

/** 
 * @param {number} x
 * @return {void}
 */
MyQueue.prototype.push = function(x) {
    this._en.push(x);
};

/**
 * @return {number}
 */
MyQueue.prototype.pop = function() {
    if(this._de.length == 0) {
        this._mv();
    }
    return this._de.pop();
};
MyQueue.prototype._mv = function() {
    while(this._en.length) {
        this._de.push(this._en.pop())
    }
    console.log(this._de, this._en)
}

/**
 * @return {number}
 */
MyQueue.prototype.peek = function() {
    const result = this.pop();
    this._de.push(result);
    return result;
};

/**
 * @return {boolean}
 */
MyQueue.prototype.empty = function() {
    return this._de.length === 0 && this._en.length === 0;
};

2.手写题：https://bigfrontend.dev/zh/problem/the-angle-between-hour-hand-and-minute-hand-of-a-clock

/**
 * @param {string} time
 * @returns {number} 
 */
const MINUTE = 60;
const HOUR = 12;
const ANGLE = 360;
const EVERY_HOUR_ANGLE = 360 / 12;
function angle(time) {
  // your code here
  // 分针的角度
  // 12:15
  const [h, m] = time.split(":");
  const ma = Math.abs(m / MINUTE * ANGLE);
  const hour = h % HOUR;
  const ha = (hour + m / MINUTE) * EVERY_HOUR_ANGLE ;
  const angle = Math.abs(ma - ha);
  return Math.round(Math.min(angle, ANGLE - angle))
}
const time = "01:00";
console.log(angle(time));

Day6

1.算法： https://leetcode-cn.com/problems/4sum/

var fourSum = function(nums, target) {
    const result = [];
    if (nums.length < 4) {
        return result;
    }
    nums.sort((x, y) => x - y);
    const length = nums.length;
    for (let i = 0; i < length - 3; i++) {
        // 排除重复项
        if (i > 0 && nums[i] === nums[i - 1]) {
            continue;
        }
        //  后面的元素都不符合，提前跳出循环
        if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
            break;
        }
        if (nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {
            continue;
        }
        for (let j = i + 1; j < length - 2; j++) {
            if (j > i + 1 && nums[j] === nums[j - 1]) {
                continue;
            }
            if (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                break;
            }
            if (nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {
                continue;
            }
            let left = j + 1, right = length - 1;
            while (left < right) {
                const sum = nums[i] + nums[j] + nums[left] + nums[right];
                if (sum === target) {
                    result.push([nums[i], nums[j], nums[left], nums[right]]);
                    while (left < right && nums[left] === nums[left + 1]) {
                        left++;
                    }
                    left++;
                    while (left < right && nums[right] === nums[right - 1]) {
                        right--;
                    }
                    right--;
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
    }
    return result;
};

2.手写map： https://bigfrontend.dev/zh/problem/implement-Array-prototype-map

Array.prototype.myMap = function(...args) {
  // your code here
  const array = this;
  const [cb,bindThis = null] = args;
  if(typeof cb !== "function") {
    throw Error("invalid function");
  }
  const result = [];
  array.forEach((item,index) => {
    result[index] = cb.call(bindThis,item, index,that);
  })
  return result
}

Day7

算法： https://leetcode-cn.com/problems/minimum-operations-to-reduce-x-to-zero/

/**
 * @param {number[]} nums
 * @param {number} x
 * @return {number}
 */
var minOperations = function(nums, x) {
    // 核心：问题等价于求解最长连续子数组，和为sum-x
    let sum = 0;
    const n = nums.length;
    for (let i=0;i<n;i++) {
            sum += nums[i];
        }
       const target = sum - x    
        if (target<0){
            return -1
        }
        
        let left = right = 0;
        // 最长连续数组【和为target】
        let ans = -1;
        // 滑动窗口中的数字总和  
        let cnt = 0       
        while( right<n) {
            cnt += nums[right]
            while(cnt>target) {
                 cnt -= nums[left]
                left += 1
            }
            
            if (cnt == target) 
                ans = Math.max(ans, right-left+1)
            
            right += 1
        }
            
        return  ans == -1 ? -1 :   n-ans
}

手写：https://bigfrontend.dev/zh/problem/lodash-set

function set(obj, path, value) {
  const pathList = Array.isArray(path) ? path : path.replace("[",".").replace("]","").split(".");
  const length = pathList.length;
  let temp = obj;
  pathList.forEach((key, index) => {
    if(index === length -1) {
      temp[key] = value;
    } else {
      if(!(key in temp)) {
        const next = pathList[index+1];
        temp[key] = Number(next) + "" === next ? [] : {};
      }
      temp = temp[key];
    }
  })
}

Day8

算法： https://leetcode-cn.com/problems/swap-nodes-in-pairs/

// 迭代
var swapPairs = function(head) {
    if(!head || !head.next) {
        return head;
    }
    const newHead = head.next;
    let prev = null;
    let next = null;
    let cur = head;
    while(cur && cur.next) {
        next = cur.next.next;
        cur.next.next = cur;
        if(prev) {
            prev.next = cur.next;
        }
        cur.next = next;
        prev = cur;
        cur = next;
    }
    return newHead;
};
// 递归
var swapPairs = function(head) {
    if(!head || !head.next) {
        return head;
    }
    const swapedList = swapPairs(head.next.next);
    newHead = head.next;
    head.next = swapedList;
    newHead.next = head;
    return newHead;
    
}

手写：https://bigfrontend.dev/zh/problem/add-comma-to-number

// toLocaleString的api用的比较少
function addComma(num) {
  const str = String(num);
  const numList = str.split('.');
  const numF = numList.length > 1 ? '.' + numList[1] : '';
  return Number(numList[0]).toLocaleString() + numF;
}
// 正则 
function addComma(num) {
  const [number, fraction] = num.toString().split('.');
  const regex = /(\d)(?=(\d{3})+$)/gism;
  const commas = number.replace(regex, (_, $1) => `${$1},`);
  return `${commas}${fraction ? '.' + fraction : ''}`;
}
// 前两种方法都是弱项
// 注意点： const intPart = Math.trunc(num);
//   const decimalPart = num  % 1; // num - intPart这两种方式浮点数的问题
function addComma(num) {
  let result = "";
  const [int, other] = String(num).split(".");
  const isNeg = num < 0;
  const count = int.replace("-", "");
  const len = count.length;
  let i = len -1;
  let counter = 0;
  while(i>=0) {
    if( i !== 0 && counter &&  counter % 3 === 0) {
    console.log("---",result);
        result = ',' + result;
    } 
    result = count[i] + result;
    counter++;
    i--;
  }
  return `${isNeg ? '-' : ''}${result}${other ? '.' + other : ''}`;
  
}

Day9

4月5日：算法题的判断条件不确定，解题时间超时

手写： https://bigfrontend.dev/zh/problem/implement-BigInt-subtraction

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
function subtract(num1, num2) {
  // your code here
  const m= num1.length;
  const n = num2.length;
  let i = m - 1;
  let j = n -1;
  let carryin = 0;
  let res = "";
  while(i>=0 && j >=0) {
    let p = num1[i--] - num2[j--] + carryin;
    if(p < 0) {
      p = p + 10;
      carryin = -1;
    } else {
      carryin = 0;
    }
    res = p + res;
   
  }
  return res.replace(/^0*/, "") || '0';
}

算法: https://leetcode-cn.com/problems/linked-list-cycle-ii/

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function(head) {
    if(!head || !head.next) {
        return null;
    }
    let slow = head;
    let fast = head;
    // fast is null, not cycle list
    while(fast && fast.next) {
        fast = fast.next.next;
        slow = slow.next;
        if(fast === slow) {
            fast = head;
            while(slow !== fast) {
                slow = slow.next;
                fast = fast.next;
            }
            return fast;
        }
    }
    return null;
};

Day10

1.算法题：https://leetcode-cn.com/problems/max-chunks-to-make-sorted-ii/

/**
 * @param {number[]} arr
 * @return {number}
 */
var maxChunksToSorted = function(arr) {
    const stack = [];
    for(let i = 0; i<arr.length;i++) {
        const cur = arr[i];
        if(stack.length && cur < stack[stack.length-1]) {
            const head = stack.pop();
            while(stack.length && stack[stack.length-1] > cur) {
                stack.pop();
            }
            stack.push(head);
        } else {
            stack.push(cur);
        }
    }
    return stack.length;
};

2.手写题：https://bigfrontend.dev/zh/problem/implement-your-own-URLSearchParams

 class MyURLSearchParams {
   /**
    * @params {string} init
    */
   constructor(init) {
    //  用来保存数据顺序
    this._paramsList = [];
    // 用来遍历和查询
    this._map = new Map();
    let _initString = String(init);
    if(_initString.startsWith("?")) {
      _initString = _initString.slice(1);
    }
    const _paramsList = _initString.split("&");
      _paramsList.forEach(item => {
        const [key, value] = item.split("=");
        this.append(key, value)
      })
   }
   
   /** 
    * @params {string} name
    * @params {any} value
    */
   append(name, value) {
     const _value = String(value);
     this._paramsList.push({
       key: name,
       value:_value,
     });
     if(this._map.has(name)) {
        this._map.get(name).push(_value);
      } else {
        this._map.set(name, [_value])
      }
   }
   
   /**
    * @params {string} name
    */
   delete(name) {
     this._paramsList = this._paramsList.filter(({key}) => {
       return key !== name
     });
     this._map.delete(name);
   }
   
   /**
    * @returns {Iterator} 
    */
   * entries() {
      for(let item of this._paramsList) {
        const {key, value} = item;
        yield [key, value]
      }
   }
   
   /**
    * @param {(value, key) => void} callback
    */
   forEach(callback) {
     this._paramsList.forEach(({key, value}) => {
       callback(value, key)
     })
   }
   
   /**
    * @param {string} name
    * returns the first value of the name
    */
   get(name) {
     if(this.has(name)) {
       return this._map.get(name)[0]
     } 
     return null;
   }
   
   /**
    * @param {string} name
    * @return {string[]}
    * returns the value list of the name
    */
   getAll(name) {
     if(this.has(name)) {
       return this._map.get(name)
     }
     return [];
   }
   
   /**
    * @params {string} name
    * @return {boolean}
    */
   has(name) {
     return this._map.has(name)
   }
   
   /**
    * @return {Iterator}
    */
    * keys() {
      for(let item of this._paramsList) {
        yield item.key
      }
   }
   
   /**
    * @param {string} name
    * @param {any} value
    */
   set(name, value) {
     const _value = String(value);
     this._map.set(name, [_value]);
     let flag = false;
     this._paramsList = this._paramsList.reduce((acc,item ) => {
       if((item.key === name && !flag) ) {
           acc.push({
             key: name, 
             _value,
           })
       }
       if(item.key !== name) {
         acc.push(item)
       }
       return acc;
     }, []);
   }
   
   // sor all key/value pairs based on the keys
   sort() {
     this._paramsList.sort(({key: k1}, {key: k2}) => {
       return k1 < k2 ? -1 : 0
     })
   }
   
   /**
    * @return {string}
    */
   toString() {
     return this._paramsList.map(({key, value}) => {
       return `${key}=${value}`
     }).join("&");
   }
   
   /**
    * @return {Iterator} values
    */
   *values() {
    for(let item of this._paramsList) {
        yield item.value
      }
   }
 }

Day11

算法题：https://leetcode-cn.com/problems/rotate-list/

//  时间复杂度：O(n)
// 空间复杂度：O(1)
var rotateRight = function(head, k) {
    if(!head) {
        return head;
    }
    const length = getListLength(head);
    k = k % length;
    let slow = head;
    let fast = head;
    while(k) {
        fast = fast.next;
        k--;
    }
    while(fast && fast.next) {
        slow = slow.next;
        prev = fast;
        fast = fast.next;
    }
    fast.next = head;
    const newHead = slow.next;
    slow.next = null
    return newHead;

};
function getListLength(head) {
    let count = 0;
    while(head) {
        count++;
        head = head.next;
    }
    return count;
}

手写：请实现一个cacheRequest(url,callback)请求缓存方法，保证当使用ajax时，对于同一个API实际在网络层只发出一次请求以节省网络流量，假设已存在request底层方法用于封装ajax请求，调用格式为： request(url, data() => {}).比如如下

function init() {
  const map = new Map();
  return function( url,cb) {
    const isExited = map.has(url);
    if(!isExited) {
      const state = {
        queue: [cb],
        status: "pending",
        result: null,
      };
      map.set(url, state);
      asyncRequest(url).then(res => {
        state.result = res;
        state.status = "fulfilled";
        state.queue.forEach(cb => {
          cb(res);
        });
        state.queue = [];
      })
    } else if(map.get(url).status == "pending") {
      map.get(url).queue.push(cb);  
    } else {
      cb(map.get(url).result)
    }
  }
}
function asyncReqest(url) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      const random = Math.random();
      resolve({
        url,
        random,
      })
    });
  });
}
const createRequest = init();
// a.js
cacheRequest('/user', data => {
   console.log('我是从A中请求的user，数据为' + JSON.stringify(data));
})
// b.js
cacheRequest('/user', data => {
   console.log('我是从B中请求的user，数据为' + JSON.stringify(data));
})

boyan1010 / everyday-al

everyday-algorithms

Day1

Day2

Day3

Day4

Day5

Day6

Day7

Day8

Day9

Day10

Day11

About