统计匹配检索规则的物品数量

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1773. 统计匹配检索规则的物品数量 - 力扣(LeetCode)

给你一个数组 items ,其中 items[i] = [typei, colori, namei] ,描述第 i 件物品的类型、颜色以及名称。

另给你一条由两个字符串 ruleKeyruleValue 表示的检索规则。

如果第 i 件物品能满足下述条件之一,则认为该物品与给定的检索规则 匹配

  • ruleKey == "type"ruleValue == typei
  • ruleKey == "color"ruleValue == colori
  • ruleKey == "name"ruleValue == namei

统计并返回 匹配检索规则的物品数量

示例 1:

输入: items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver"
输出: 1
解释: 只有一件物品匹配检索规则,这件物品是 ["computer","silver","lenovo"] 。

示例 2:

输入: items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone"
输出: 2
解释: 只有两件物品匹配检索规则,这两件物品分别是 ["phone","blue","pixel"] 和 ["phone","gold","iphone"] 。注意,["computer","silver","phone"] 未匹配检索规则。

提示:

  • 1 <= items.length <= 10^4
  • 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
  • ruleKey 等于 "type""color""name"
  • 所有字符串仅由小写字母组成

思路

这道题很简单,先通过ruleKey确定物品属性下标,再遍历物品列表,比较对应数据项和ruleValue是否相等,相等计数器加一,最后返回计数结果就是要求的值,如解法一。
也可通过Array.property.filter求解,只需三行代码。如解法二。

解题

解法一

/**
 * @param {string[][]} items
 * @param {string} ruleKey
 * @param {string} ruleValue
 * @return {number}
 */
var countMatches = function (items, ruleKey, ruleValue) {
  let idx = ruleKey === "type" ? 0 : ruleKey === "color" ? 1 : 2;
  let count = 0;
  for (let i = 0; i < items.length; i++) {
    if (items[i][idx] === ruleValue) {
      count++;
    }
  }
  return count
};

解法二

/**
 * @param {string[][]} items
 * @param {string} ruleKey
 * @param {string} ruleValue
 * @return {number}
 */
var countMatches = function (items, ruleKey, ruleValue) {
  let idx = ruleKey === "type" ? 0 : ruleKey === "color" ? 1 : 2;
  let filter = (item) => item[idx] === ruleValue;
  return items.filter(filter).length;
};