LeetCode 1773. 统计匹配检索规则的物品数量

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)