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1773. 统计匹配检索规则的物品数量#

问题描述#

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

解题思路#

遍历判断就行。

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class Solution:
    def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
        ans = k = 0

        if ruleKey == "color":
            k = 1
        elif ruleKey == "name":
            k = 2

        for item in items:
            if item[k] == ruleValue:
                ans += 1

        return ans

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class Solution {
public:
    int countMatches(vector<vector<string>>& items, string ruleKey, string ruleValue) {
        int ans = 0;
        int k = 0;

        if (ruleKey == "color") k = 1;
        else if (ruleKey == "name") k = 2;

        for (auto &item : items) {
            if (item[k] == ruleValue) ++ans;
        }

        return ans;
    }
};

时间复杂度：\(\mathcal{O}(n)\)
空间复杂度：\(\mathcal{O}(n)\)

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