计匹配检索规则的物品数量 || 刷题打卡

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题目

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 ruleKey 和 ruleValue 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" 且 ruleValue == typei 。
  • ruleKey == "color" 且 ruleValue == colori 。
  • ruleKey == "name" 且 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"]["phone","gold","iphone"] 。注意,["computer","silver","phone"] 未匹配检索规则。

提示:

  • 1 <= items.length <= 104
  • 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
  • ruleKey 等于 "type"、"color" 或 "name"
  • 所有字符串仅由小写字母组成

解题思路

这是一道简单的匹配题,我们首先判断传进来的值是类型,是颜色还是名称,然后在数组中去遍历,去记录与之匹配成功的次数,最后累计起来返回出去,既然需要累计,那我们就用reduce这个方法来做我们遍历的方法

reduce参数(为不懂reduce参数做一下解释)

WechatIMG244715.jpeg

解题

/**
 * @param {string[][]} items
 * @param {string} ruleKey
 * @param {string} ruleValue
 * @return {number}
 */
var countMatches = function(items, ruleKey, ruleValue) {
    // 首先我们知道我们需要匹配的顺序都是依次排列的,这样我们就可以通过[]取值的方法来做匹配
    const map = {
        type: 0,
        color: 1,
        name: 2,
    }
    // 
    return items.reduce((acc, curr) => {
        // 匹配成功,个数加1
        if(curr[map[ruleKey]] === ruleValue) {
            acc ++
        }
        return acc;
    },0) // 默认传递0
};

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