22. Generate Parentheses

2020-05-23

原题：22. Generate Parentheses

source code

// Runtime: 68 ms, faster than 23.67% of JavaScript online submissions for Generate Parentheses.
// Memory Usage: 37.1 MB, less than 7.69% of JavaScript online submissions for Generate Parentheses.
var generateParenthesis = function (n) {
  const result = [];
  const helper = function (left, right, str) {
    //left:左括号还剩几个，right:右括号还剩几个，str:当前结果子串
    if (left === 0 && right === 0) {
      result.push(str);
    }
    if (left > 0) {
      //只要当前子串合法，左括号可以随意加
      helper(left - 1, right, str + "(");
    }
    if (right > left) {
      //如果当前子串合法，子串中的左括号一定比右括号多
      helper(left, right - 1, str + ")");
    }
  };

  helper(n, n, "");

  return result;
};

test cases

test("test1", () => {
  expect(generateParenthesis(1).sort()).toEqual(["()"].sort());
});

test("test2", () => {
  expect(generateParenthesis(2).sort()).toEqual(["()()", "(())"].sort());
});

test("test3", () => {
  expect(generateParenthesis(3).sort()).toEqual(
    ["((()))", "(()())", "(())()", "()(())", "()()()"].sort()
  );
});

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111. Minimum Depth of Binary Tree

2020-05-23

原题：111. Minimum Depth of Binary Tree

方法一

思路：DFS
时间复杂度：O(n)

// Runtime: 80 ms, faster than 17.32% of JavaScript online submissions for Minimum Depth of Binary Tree.
// Memory Usage: 38.4 MB, less than 20.00% of JavaScript online submissions for Minimum Depth of Binary Tree.
var minDepth = function (root) {
  if (!root) {
    return 0;
  }

  if (root.left && root.right) {
    return 1 + Math.min(minDepth(root.left), minDepth(root.right));
  } else if (root.left) {
    return 1 + minDepth(root.left);
  } else if (root.right) {
    return 1 + minDepth(root.right);
  } else {
    return 1;
  }
};

方法二

思路：BFS
时间复杂度：O(n)

// Runtime: 72 ms, faster than 21.72% of JavaScript online submissions for Minimum Depth of Binary Tree.
// Memory Usage: 40.6 MB, less than 20.00% of JavaScript online submissions for Minimum Depth of Binary Tree.
var minDepth = function (root) {
  if (!root) {
    return 0;
  }

  const queue = [[root, 0]];
  let minLevel = Infinity;
  while (queue.length) {
    const [node, level] = queue.shift();
    !node.left && !node.right && (minLevel = Math.min(minLevel, level));
    node.left && queue.push([node.left, level + 1]);
    node.right && queue.push([node.right, level + 1]);
  }

  return minLevel + 1;
};

test cases

test("test1", () => {
  expect(minDepth(makeTree([3, 9, 20, null, null, 15, 7]))).toEqual(2);
});

test("test2", () => {
  expect(minDepth(makeTree([]))).toEqual(0);
});

test("test3", () => {
  expect(minDepth(makeTree([1]))).toEqual(1);
});

test("test4", () => {
  expect(minDepth(makeTree([1, 2]))).toEqual(2);
});

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104. Maximum Depth of Binary Tree

2020-05-23

原题：104. Maximum Depth of Binary Tree

方法一

思路：DFS + recursive
时间复杂度：O(n)

// Runtime: 68 ms, faster than 33.83% of JavaScript online submissions for Maximum Depth of Binary Tree.
// Memory Usage: 38.5 MB, less than 6.25% of JavaScript online submissions for Maximum Depth of Binary Tree.
var maxDepth = function (root) {
  if (!root) {
    return 0;
  }

  return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
};

方法二

思路：BFS + iterative
时间复杂度：O(n)

// Runtime: 88 ms, faster than 15.71% of JavaScript online submissions for Maximum Depth of Binary Tree.
// Memory Usage: 40.1 MB, less than 6.25% of JavaScript online submissions for Maximum Depth of Binary Tree.
var maxDepth = function (root) {
  if (!root) {
    return 0;
  }

  const queue = [[root, 0]];
  let maxLevel = 0;
  while (queue.length) {
    const [node, level] = queue.shift();
    maxLevel = Math.max(maxLevel, level);
    node.left && queue.push([node.left, level + 1]);
    node.right && queue.push([node.right, level + 1]);
  }

  return maxLevel + 1;
};

test cases

test("test1", () => {
  expect(maxDepth(makeTree([3, 9, 20, null, null, 15, 7]))).toEqual(3);
});

test("test2", () => {
  expect(maxDepth(makeTree([]))).toEqual(0);
});

test("test3", () => {
  expect(maxDepth(makeTree([1]))).toEqual(1);
});

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102. Binary Tree Level Order Traversal

2020-05-23

原题：102. Binary Tree Level Order Traversal

方法一

思路：BFS
时间复杂度：O(n)

// Runtime: 72 ms, faster than 19.61% of JavaScript online submissions for Binary Tree Level Order Traversal.
// Memory Usage: 35.6 MB, less than 5.88% of JavaScript online submissions for Binary Tree Level Order Traversal.
var levelOrder = function (root) {
  if (!root) {
    return [];
  }

  const queue = [[root, 0]];
  const result = [];
  while (queue.length) {
    const [node, index] = queue.shift();
    result[index] = result[index] || [];
    result[index].push(node.val);
    node.left && queue.push([node.left, index + 1]);
    node.right && queue.push([node.right, index + 1]);
  }
  
  return result;
};

方法二

思路：和方法一的思路类似，也是通过层级计数，不过通过DFS实现
时间复杂度：O(n)

// Runtime: 68 ms, faster than 21.98% of JavaScript online submissions for Binary Tree Level Order Traversal.
// Memory Usage: 35.4 MB, less than 5.88% of JavaScript online submissions for Binary Tree Level Order Traversal.
var levelOrder = function (root, result = [], level = 0) {
  if (!root) {
    return result;
  }

  result[level] = result[level] || [];
  result[level].push(root.val);
  root.left && levelOrder(root.left, result, level + 1);
  root.right && levelOrder(root.right, result, level + 1);

  return result;
};

方法三

思路：BFS + Batch Process
时间复杂度：O(n)

// Runtime: 64 ms, faster than 30.81% of JavaScript online submissions for Binary Tree Level Order Traversal.
// Memory Usage: 35.7 MB, less than 5.88% of JavaScript online submissions for Binary Tree Level Order Traversal.
var levelOrder = function (root) {
  if (!root) {
    return [];
  }
  const queue = [root];
  const result = [];
  while (queue.length) {
    let level = [];
    for (let i = 0, len = queue.length; i < len; i++) {
      //注意这里把层节点数记录了下来
      const node = queue.shift();
      level.push(node.val);
      node.left && queue.push(node.left);
      node.right && queue.push(node.right);
    }
    result.push(level);
  }

  return result;
};

test cases

test("test1", () => {
  expect(levelOrder(makeTree([3, 9, 20, null, null, 15, 7]))).toEqual([
    [3],
    [9, 20],
    [15, 7],
  ]);
});

test("test2", () => {
  expect(levelOrder(makeTree([]))).toEqual([]);
});

test("test3", () => {
  expect(levelOrder(makeTree([1]))).toEqual([[1]]);
});

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Best Time to Buy and Sell Stock

2020-05-22

121. Best Time to Buy and Sell Stock

原题：121. Best Time to Buy and Sell Stock

source code

// Runtime: 80 ms, faster than 33.60% of JavaScript online submissions for Best Time to Buy and Sell Stock.
// Memory Usage: 38.6 MB, less than 7.41% of JavaScript online submissions for Best Time to Buy and Sell Stock.
var maxProfit = function (prices) {
  let minBuy = prices.shift();
  let maxProfit = 0;
  for (let price of prices) {
    maxProfit = Math.max(maxProfit, price - minBuy);
    minBuy = Math.min(price, minBuy);
  }

  return maxProfit;
};

test cases:

test("test1", () => {
  expect(maxProfit([1, 2, 3, 4, 5])).toEqual(4);
});

test("test2", () => {
  expect(maxProfit([7, 6, 4, 3, 1])).toEqual(0);
});

test("test3", () => {
  expect(maxProfit([7, 1, 5, 3, 6, 4])).toEqual(5);
});

122. Best Time to Buy and Sell Stock II

原题：122. Best Time to Buy and Sell Stock II

这道题的思路是：因为能买卖无数次，只要能赚钱（价格比前一天高），就尽快卖出。能同一天进行买卖。

source code

// Runtime: 84 ms, faster than 27.73% of JavaScript online submissions for Best Time to Buy and Sell Stock II.
// Memory Usage: 37 MB, less than 23.81% of JavaScript online submissions for Best Time to Buy and Sell Stock II.
var maxProfit = function (prices) {
  let sum = 0;
  for (let i = 0, j = 1; j < prices.length; ) {
    if (prices[j] > prices[i]) {
      sum += prices[j] - prices[i];
    }
    i = j;
    j++;
  }

  return sum;
};

test cases

test("test1", () => {
  expect(maxProfit([7, 1, 5, 3, 6, 4])).toEqual(7);
});

test("test2", () => {
  expect(maxProfit([7, 6, 4, 3, 1])).toEqual(0);
});

test("test3", () => {
  expect(maxProfit([1, 2, 3, 4, 5])).toEqual(4);
});

123. Best Time to Buy and Sell Stock III

原题：123. Best Time to Buy and Sell Stock III

source code

// Runtime: 2064 ms, faster than 5.19% of JavaScript online submissions for Best Time to Buy and Sell Stock III.
// Memory Usage: 47.7 MB, less than 50.00% of JavaScript online submissions for Best Time to Buy and Sell Stock III.
var maxProfit = function (prices) {
  const helper = function (prices) {//helper实际上是121的代码
    let minBuy = prices.shift();
    let maxProfit = 0;
    for (let price of prices) {
      maxProfit = Math.max(maxProfit, price - minBuy);
      minBuy = Math.min(price, minBuy);
    }

    return maxProfit;
  };

  let maxProfit = 0;
  for (let i = 1; i <= prices.length; i++) {
    maxProfit = Math.max(
      maxProfit,
      helper(prices.slice(0, i)) + helper(prices.slice(i))
    );
  }

  return maxProfit;
};

test code

test("test1", () => {
  expect(maxProfit([3, 3, 5, 0, 0, 3, 1, 4])).toEqual(6);
});

test("test2", () => {
  expect(maxProfit([7, 6, 4, 3, 1])).toEqual(0);
});

test("test3", () => {
  expect(maxProfit([1, 2, 3, 4, 5])).toEqual(4);
});

test("test4", () => {
  expect(maxProfit([2, 1, 2, 0, 1])).toEqual(2);
});

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模拟es6 Class中的private/public

2020-05-22

阿里笔试中的一道题，一眼看上去挺简单，但是仔细想想有挺多地方要注意的。。

原题

请实现一个Person类，且拥有private的成员变量及函数，public的成员变量及函数

提示：参考ts实现

class Person {
  private name = '';

  public address = '';

  constructor(name: string, address: string) {
    this.name = name;
    this.address = address;
  }

  private makeWords() {
    return `My name is: ${this.name}`;
  }

  public say() {
    console.log(this.makeWords());
  }
}

//结果预期
me = new Person('alibaba', 'hangzhou');
me.say() // My name is: alibaba
me.makeWords() // Error
me.name // undefined
me.address // hangzhou

解题过程

第一眼看到这道题的时候马上想到的是最简单的实现，用约定俗成的方法，给私有变量/方法名前缀加下划线：

方法一

const Person = function(name, address) {
    this._name = name;
    this.address = address;
    // 其他类似
}

方法二

法一只是一种约定，外部还是能通过_name取到私有属性。然后想到通过Object.defineProperty或者Proxy拦截私有属性的访问，于是写了一个版本：

const Person = function (name, address) {
  const obj = {
    name,
    address,
    makeWords() {
      return `My name is: ${this._safeGet("name")}`;
    },
    say() {
      console.log(this._safeGet("makeWords")());
    },
    _safeGet(propName) {
      const propVal = Reflect.get(obj, propName);
      return typeof propVal === "function" ? propVal.bind(this) : propVal;
    },
  };

  return new Proxy(obj, {
    get(target, propName) {
      const privateProps = ["name", "makeWords"];
      if (privateProps.includes(propName)) {
        return undefined;
      } else {
        return typeof target[propName] === "function"
          ? target[propName].bind(target)
          : target[propName]; //proxy拦截方法里面的this指向Proxy
      }
    },
  });
};

这个方法能屏蔽外部对私有属性的访问，但是内部调用必须经过_safeGet，和题目要求有微妙的区别。。而且某些浏览器不支持Proxy。

方法三

相当于方法二的简单实现，用闭包屏蔽外部访问，还是没解决内部调用的问题：

const Person = function (name, address) {
  const makeWords = function () {
    return `My name is: ${name}`;//怎么改造可以使用this.name？
  };

  const say = function () {
    console.log(makeWords());//怎么改造可以使用this.makeWords？
  };

  return { address, say };
};

方法四

通过原型链屏蔽私有属性的访问，这个解法是符合题目要求的，但是是否是最优解不得而知：

const Person = function (name, address) {
  const privateObj = {};
  const publicObj = {};
  privateObj.__proto__ = publicObj;

  const makeWords = function () {
    return `My name is: ${this.name}`;
  };

  const say = function () {
    console.log(this.makeWords());
  };

  const props = [//属性配置元组
    ["name", name, true], //propName, propValue, isPrivate
    ["address", address, false],
    ["makeWords", makeWords, true],
    ["say", say, false],
  ];

  for (let [propName, propValue, isPrivate] of props) {
    if (typeof propValue === "function") {
      propValue = propValue.bind(privateObj);
    }
    (isPrivate ? privateObj : publicObj)[propName] = propValue;
  }

  return publicObj;
};

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19. Remove Nth Node From End of List

2020-05-22

原题：19. Remove Nth Node From End of List

这道题考察的是滑动窗口。

解法一

时间复杂度：O(n)

// Runtime: 76 ms, faster than 14.67% of JavaScript online submissions for Remove Nth Node From End of List.
// Memory Usage: 34.8 MB, less than 13.64% of JavaScript online submissions for Remove Nth Node From End of List.
var removeNthFromEnd = function (head, n) {
  const queue = [];
  const enqueue = function (node) {
    if (queue.length >= n + 1) {
      queue.shift();
    }
    queue.push(node);
  };

  const _head = new ListNode();
  _head.next = head;
  let cur = _head;
  while (cur) {
    enqueue(cur);
    cur = cur.next;
  }

  queue[0].next = queue[2] || null; //删除
  queue[1].next = null; //清理

  return _head.next;
};

时间复杂度：O(n)

比解法一优化了空间复杂度，只保留了窗口的头尾指针。

// Runtime: 96 ms, faster than 8.34% of JavaScript online submissions for Remove Nth Node From End of List.
// Memory Usage: 34.1 MB, less than 68.18% of JavaScript online submissions for Remove Nth Node From End of List.
var removeNthFromEnd = function (head, n) {
  const _head = new ListNode();
  _head.next = head;
  let left = _head;
  let right = _head;

  while (n-- >= 0) {
    right = right.next;
  }
  while (right) {
    left = left.next;
    right = right.next;
  }
  left.next = left.next.next;

  return _head.next;
};

Test Cases:

test("test1", () => {
  expect(removeNthFromEnd(makeLink([1, 2, 3, 4, 5]), 2)).toEqual(
    makeLink([1, 2, 3, 5])
  );
});

test("test2", () => {
  expect(removeNthFromEnd(makeLink([1]), 1)).toEqual(makeLink([]));
});

展开全文 >>

为contenteditable增加placeholder

2020-05-22

加了contenteditable的div默认是没有placeholder的，下面尝试用css实现一个placeholder的效果：

1	<div class="richTextBox empty" contenteditable placeholder="请输入"></div>

.richTextBox {
    border: 1px solid #eee;
    height: 100px;
}
.richTextBox.empty:before {
    position: absolute; /*强制使伪元素脱离文档流，否则ff下输入的文字会跟在placeholder后面*/
    content: attr(placeholder); /*注意这里不能用引号*/
    color: #ddd;
    cursor: text;
}

效果图：

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用纯css实现一个iOS的switch开关

2020-05-21

需求：

使用一个 div，不借助 js 能力的前提下，纯 css 实现一个类似如下图的效果：
iOS switch
hover上去的时候变化。

实现

印象中这道题在当年阿里校招的笔试上见过，但是很多年没写css了，很多知识点忘的差不多了，答得不算好。下面总结下：

.switch {
    position: relative;
    width: 100px;
    height: 30px;
    background: #3388ff;
    font-size: 0px;
    transition: 1s ease;
}
.switch:before {
    content: "";
    position: absolute;
    top: 0;
    left: 0;
    display: inline-block;
    width: 30px;
    height: 30px;
    border-radius: 15px;
    background: #eee;
    transition: 1s ease;
}
.switch:hover {
    background: #eee;
}
.switch:hover:before {
    background: #3388ff;
    /*注意这里要用left，因为transition是通过计算前面属性的变化做补间动画的。笔试的时候写成了right:0；导致滑块瞬移无动画。。*/
    left: 70px;
}

1	<div class="switch"></div>

关于transition这篇文章总结得不错，可以看看：
深入理解CSS过渡transition。

拓展

和checkbox结合实现点击动画

.switch {
    display: inline-block;
    font-size: 0;
}
.switch input {
    display: none;
}
.switch b {
    position: relative;
    display: inline-block;
    width: 100px;
    height: 30px;
    box-sizing: border-box;
    background: #3388ff;
    font-size: 0px;
    transition: 1s ease;
}
.switch b:before {
    content: "";
    position: absolute;
    display: inline-block;
    width: 30px;
    height: 30px;
    border-radius: 15px;
    background: #eee;
}
.switch input:checked ~ b {
    background: #eee;
    padding-left: 70px;
}

.switch input:checked ~ b:before {
    background: #3388ff;
}

<label class="switch">
    <input type="checkbox" />
    <b></b>
</label>

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169. Majority Element

2020-05-21

原题：169. Majority Element

解法一

思路：brute force，循环嵌套依次找出每个元素出现的次数；
时间复杂度：O(n^2)

解法二

思路：用map计数，找出出现次数>长度一半的元素
时间复杂度：O(n)

// Runtime: 120 ms, faster than 5.64% of JavaScript online submissions for Majority Element.
// Memory Usage: 38.1 MB, less than 21.43% of JavaScript online submissions for Majority Element.
var majorityElement = function (nums) {
  const map = {};
  for (let num of nums) {
    map[num] = (map[num] || 0) + 1;
    if (map[num] >= nums.length / 2) {
      return num;
    }
  }

  //题干保证众数一定存在
};

解法三

思路：先把数组排序，众数一定在中间的位置
时间复杂度：O(n*logn)

// Runtime: 84 ms, faster than 10.46% of JavaScript online submissions for Majority Element.
// Memory Usage: 37.5 MB, less than 57.14% of JavaScript online submissions for Majority Element.
var majorityElement = function (nums) {
  nums.sort();
  return nums[~~(nums.length / 2)];
};

解法四

思路：分治。把数组一分为二，找出前半段的众数以及众数出现的次数（lm, lc），后半段的众数以及众数出现的次数（rm, rc）。如果lm==lc，则众数为lm；否则比较lc和rc，返回较大的对应lm/rm。
时间复杂度：O(logn+n)

var majorityElement = function (nums, start = 0, end) {
  const findNumCount = function (num, start, end) {
    let count = 0;
    for (let i = start; i <= end; i++) {
      if (nums[i] == num) {
        count++;
      }
    }
    return count;
  };
  const helper = function (start, end) {
    if (end - start == 0) {
      return [nums[start], 1];
    }

    const midIndex = ~~((start + end) / 2);
    const [lm, lc] = helper(start, midIndex);
    const [rm, rc] = helper(midIndex + 1, end);
    if (lm == rm) {
      return [lm, lc + rc];
    } else {
      return lc > rc
        ? [lm, lc + findNumCount(lm, midIndex + 1, end)]
        : [rm, rc + findNumCount(rm, start, midIndex)];
    }
  };

  return helper(0, nums.length - 1)[0];
};

Test Cases:

test("test1", () => {
  expect(majorityElement([3, 2, 3])).toEqual(3);
});

test("test2", () => {
  expect(majorityElement([2, 2, 1, 1, 1, 2, 2])).toEqual(2);
});

test("test3", () => {
  expect(
    majorityElement([
      345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 543, 345, 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543, 345
    ])
  ).toEqual(345);
});

展开全文 >>

source code

test cases

方法一

方法二

test cases

方法一

方法二

test cases

方法一

方法二

方法三

test cases

121. Best Time to Buy and Sell Stock

source code

test cases:

122. Best Time to Buy and Sell Stock II

source code

test cases

123. Best Time to Buy and Sell Stock III

source code

test code

原题

解题过程

方法一

方法二

方法三

方法四

解法一

解法二（Recommend）

Test Cases:

需求：

实现

拓展

解法一

解法二

解法三

解法四

Test Cases: