class Solution {
public:
int countMatches(vector<vector<string>>& items, string ruleKey, string ruleValue) {
}
};
1773. 统计匹配检索规则的物品数量
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 <= 1041 <= typei.length, colori.length, namei.length, ruleValue.length <= 10ruleKey 等于 "type"、"color" 或 "name"原站题解
rust 解法, 执行用时: 8 ms, 内存消耗: 4 MB, 提交时间: 2023-09-12 17:57:10
/*
impl Solution {
pub fn count_matches(items: Vec<Vec<String>>, rule_key: String, rule_value: String) -> i32 {
items.iter().fold(0, |cnt, item| (item[match &rule_key as &str { "type" => 0, "color" => 1, "name" => 2, _ => 3}] == rule_value) as i32 + cnt)
}
}
*/
impl Solution {
pub fn count_matches(items: Vec<Vec<String>>, rule_key: String, rule_value: String) -> i32 {
items.iter().fold(0, |cnt, item| match rule_key.as_str() {
"type" => if item[0] == rule_value { cnt + 1 } else { cnt }
"color" => if item[1] == rule_value { cnt + 1 } else { cnt }
"name" => if item[2] == rule_value { cnt + 1 } else { cnt }
_ => cnt
})
}
}
/*
impl Solution {
pub fn count_matches(items: Vec<Vec<String>>, rule_key: String, rule_value: String) -> i32 {
let i = match rule_key.as_str() {
"type" => 0,
"color" => 1,
"name" => 2,
_ => panic!(),
};
items
.into_iter()
.filter(|item| item[i] == rule_value)
.count() as i32
}
}
*/
javascript 解法, 执行用时: 72 ms, 内存消耗: 44.9 MB, 提交时间: 2023-09-12 17:55:51
/**
* @param {string[][]} items
* @param {string} ruleKey
* @param {string} ruleValue
* @return {number}
*/
var countMatches = function(items, ruleKey, ruleValue) {
const index = {"type":0, "color":1, "name":2}[ruleKey];
let res = 0;
for (const item of items) {
if (item[index] === ruleValue) {
res++;
}
}
return res;
};
java 解法, 执行用时: 4 ms, 内存消耗: 46.6 MB, 提交时间: 2023-09-12 17:55:21
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;
}
}
python3 解法, 执行用时: 68 ms, 内存消耗: 21.2 MB, 提交时间: 2023-09-12 17:55:01
class Solution:
def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
index = {"type": 0, "color": 1, "name": 2}[ruleKey]
return sum(item[index] == ruleValue for item in items)
golang 解法, 执行用时: 40 ms, 内存消耗: 6.9 MB, 提交时间: 2021-06-09 10:47:38
func countMatches(items [][]string, ruleKey string, ruleValue string) int {
ans := 0
i := 0
if ruleKey == "color" {
i = 1
} else if ruleKey == "name" {
i = 2
}
for _, item := range items {
if item[i] == ruleValue {
ans++
}
}
return ans
}