carloscn
commented
1 year ago 问题解析
pub fn count_matches(items: Vec<Vec<&str>>, rule_key: &str, rule_value: &str) -> i32
{
if items.is_empty() || rule_key.is_empty() || rule_value.is_empty() {
return 0;
}
let index:usize;
if rule_key == "type" {
index = 0;
} else if rule_key == "color" {
index = 1;
} else if rule_key == "name" {
index = 2;
} else {
return 0;
}
let mut ret:i32 = 0;
for e in &items {
if (*e)[index] == rule_value {
ret += 1;
}
}
return ret;
}
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