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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"] 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
解题思路
class Solution: def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int: type = { "type": 0, "color": 1, "name": 2 } index = type[ruleKey] resList = [i[index] for i in items] return resList.count(ruleValue) if __name__ == '__main__': # items = [["phone", "blue", "pixel"], ["computer", "silver", "lenovo"], # ["phone", "gold", "iphone"]] # ruleKey = "color" # ruleValue = "silver" items = [["phone", "blue", "pixel"], ["computer", "silver", "phone"], ["phone", "gold", "iphone"]] ruleKey = "type" ruleValue = "phone" ret = Solution().countMatches(items, ruleKey, ruleValue) print(ret)
