题目描述

给你一个数组 items，其中 items[i] = [type_i, color_i, name_i]，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 == type_i。
• ruleKey == "color" 且 ruleValue == color_i。
• ruleKey == "name" 且 ruleValue == name_i。

统计并返回 匹配检索规则的物品数量。

样例

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"] 。

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。

限制

• 1 <= items.length <= 10^4
• 1 <= type_i.length, color_i.length, name_i.length, ruleValue.length <= 10
• ruleKey 等于 "type"、"color" 或 "name"。
• 所有字符串仅由小写字母组成。

算法

(模拟) $O(n)$

1. 按照题目描述模拟即可。

时间复杂度

• 遍历数组一次，故时间复杂度为 $O(n)$。

空间复杂度

• 仅需要常数的额外空间。

C++ 代码

class Solution {
public:
    int countMatches(vector<vector<string>>& items, string ruleKey, string ruleValue) {
        int res = 0;
        for (const auto &item : items)
            if (ruleKey == "type" && item[0] == ruleValue) res++;
            else if (ruleKey == "color" && item[1] == ruleValue) res++;
            else if (ruleKey == "name" && item[2] == ruleValue) res++;

        return res;
    }
};