1773. 统计匹配检索规则的物品数量（难度：简单）

题目链接：https://leetcode.cn/problems/count-items-matching-a-rule/

题目描述：

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"] 和 ["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

解法：模拟

先根据ruleKey来获取需要匹配的物品特性下标，在遍历物品列表，判断其对应特性是否满足条件。

代码：

class Solution {
    public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
        int key = -1;
        if("type".equals(ruleKey)) {
            key = 0;
        } else if("color".equals(ruleKey)) {
            key = 1;
        } else {
            key = 2;
        }
        int result = 0;
        for(int i = 0;i<items.size();i++) {
            if(ruleValue.equals(items.get(i).get(key))) {
                result++;
            }
        }
        return result;
    }
}

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