1773. 统计匹配检索规则的物品数量
难度:简单
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 <= 10^4
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
个人题解
思路:
通过key确定索引,再逐个判断value即可
java
class Solution {
public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
int count = 0;
int keyIndex = "type".equals(ruleKey) ? 0 : "color".equals(ruleKey) ? 1 : 2;
for (List<String> item : items) {
if (item.get(keyIndex).equals(ruleValue)) {
count++;
}
}
return count;
}
}
官方题解
方法一:模拟
可以利用哈希表把输入 ruleKey 转换为 item[i] 的下标,然后再遍历一遍 items ,找出符合条件的物品数量。
java
class Solution {
public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
int index = new HashMap<String, Integer>() {{
put("type", 0);
put("color", 1);
put("name", 2);
}}.get(ruleKey);
int res = 0;
for (List<String> item : items) {
if (item.get(index).equals(ruleValue)) {
res++;
}
}
return res;
}
}
作者:力扣官方题解
来源:力扣(LeetCode)
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