30天刷题挑战（十九）

2462. 雇佣 K 位工人的总代价

思路

使用两个优先队列存放 costs 左右两边的数据，每次判断左右队列头的大小，将小的数出列加到总代价上，一直到雇佣数为 0，返回总代价。

代码

js 复制代码

/**
 * @param {number[]} costs
 * @param {number} k
 * @param {number} candidates
 * @return {number}
 */
var totalCost = function(costs, k, candidates) {
  const left = new MinPriorityQueue();
  const right = new MinPriorityQueue();
  

  let cost = 0
  let i = 0
  let j = costs.length -1

  while (k > 0) {
    while(i <= j && left.size() < candidates) {
      left.enqueue(costs[i++])
    }

    while(i <= j && right.size() < candidates) {
      right.enqueue(costs[j--])
    }

    const min1 = left.size() > 0 ? left.front().element : Infinity
    const min2 = right.size() > 0 ? right.front().element : Infinity
   
    if (min1 <= min2) {
      cost += left.dequeue().element
    } else {
      cost += right.dequeue().element
    }
    k--
  }

  return cost
};

374. 猜数字大小

思路

二分查找，每次寻找目标数在左还是右区间，当 guess(mid) 等于 0 时返回 left

代码

js 复制代码

/** 
 * Forward declaration of guess API.
 * @param {number} num   your guess
 * @return 	     -1 if num is higher than the picked number
 *			      1 if num is lower than the picked number
 *               otherwise return 0
 * var guess = function(num) {}
 */

/**
 * @param {number} n
 * @return {number}
 */
var guessNumber = function(n) {
  let left = 1
  let right = n
  
  while(left  < right) {
    const mid = Math.floor(left + (right - left) / 2)

    if (guess(mid) <= 0) {
      right = mid
    } else {
      left = mid + 1
    }
  }

  return left 
};

2300. 咒语和药水的成功对数

思路

先将 potions 从小到大排序，遍历 spells 数组，每次通过二分查找找到 potions 数组全部强度有效的下标 left，potions.length - left 表示有效数字的个数，添加到结果中。

代码

js 复制代码

/**
 * @param {number[]} spells
 * @param {number[]} potions
 * @param {number} success
 * @return {number[]}
 */
var successfulPairs = function(spells, potions, success) {
  potions.sort((a, b) => a - b)
  let res = []

  for(let i = 0; i < spells.length; i++) {
    let mid
    let left = 0
    let right = potions.length

    while(left < right) {
      mid = Math.floor(left + (right - left) / 2)
      if (potions[mid] * spells[i] >= success) {
        right = mid
      } else {
        left = mid + 1
      }
    }
    
    res.push(potions.length - left)
  }

  return res
};

本文完，感谢阅读🌹