算法题

  • 链接：https://leetcode-cn.com/problems/find-bottom-left-tree-value/submissions/
  • 题目：
    • 给定一个二叉树的 根节点 root，请找出该二叉树的 最底层 最左边 节点的值。
    • 假设二叉树中至少有一个节点。

  • 解题

    1. var findBottomLeftValue = function(root) {
    2.   let maxLen = 0;
    3.   let res = null;
    4.   const getLeftNode=function(node,curLen) {
    5.       if (!node.left && !node.right) {
    6.           if (curLen>maxLen) {
    7.               maxLen = curLen
    8.               res = node.val
    9.           }
    10.       }
    11.       node.left && getLeftNode(node.left,curLen+1)
    12.       node.right && getLeftNode(node.right,curLen+1)
    13.   }
    14.   getLeftNode(root ,1)
    15.   return res
    16. };

  • 知道用递推，一写就废，自己写的很拧巴，待遇加强递归的终结条件

    手写题

  • 题目 ```javascript 实现一个LazyMan，可以按照以下方式调用: LazyMan(“Hank”)输出: Hi! This is Hank!

    LazyMan(“Hank”).sleep(10).eat(“dinner”)输出 Hi! This is Hank! //等待10秒.. Wake up after 10 Eat dinner~

    LazyMan(“Hank”).eat(“dinner”).eat(“supper”)输出 Hi This is Hank! Eat dinner~ Eat supper~ LazyMan(“Hank”).eat(“supper”).sleepFirst(5)输出 //等待5秒 Wake up after 5 Hi This is Hank! Eat supper

  2. - 解题
  3. ```javascript
  4.   function delay(time) {
  5.       return new Promise(resolve => {
  6.         setTimeout(() => {
  7.           resolve()
  8.         }, time);
  9.       })
  10.     }
  11.     function LazyMan(value) {
  12.       const tasks = [{
  13.         type: 'immediately',
  14.         value: ` Hi! This is ${value}!`
  15.       }]
  16.       let actions = {
  17.         eat(food) {
  18.           tasks.push({
  19.             type: 'immediately',
  20.             text: ` Eat ${food}!`,
  21.             value:food
  22.           })
  23.           return this
  24.         },
  25.         sleep(time) {
  26.           tasks.push({
  27.             type: 'delay',
  28.             text: ` Wake up after ${time}!`,
  29.             value:time
  30.           })
  31.           return this
  32.         },
  33.         sleepFirst(time) {
  34.           tasks.unshift({
  35.             type: 'delay',
  36.             text: ` Wake up after ${time}!`,
  37.             value:time
  38.           })
  39.           return this
  40.         }
  41.       }
  42.       async function execFn (){
  43.         for (let item of tasks){
  44.           if (item.type ==='delay') {
  45.             await delay(item.value)
  46.             console.log(item.text)
  47.           } else {
  48.             console.log(item.value)
  49.           }
  50.         }
  51.       }
  52.       setTimeout(execFn);
  53.       return actions
  54.     }
  57.     LazyMan("Hank").eat("dinner").sleepFirst(1000);