力扣刷题笔记--1773. 统计匹配检索规则的物品数量

C_H

发布于 2022-11-15 15:00:10

2980

发布于 2022-11-15 15:00:10

文章被收录于专栏：笔记c

题目描述：

简单题

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

代码

python代码

class Solution:
    def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
        x=['type','color','name']
        count=0
        for i in range(3):
            if ruleKey==x[i]:
                n=i
                break
        for i in range(len(items)):
            if items[i][n]==ruleValue:
                count=count+1
        return count

时间52ms

另一种python代码：

class Solution:
    def countMatches(self, items: List[List[str]], ruleKey: str, ruleValue: str) -> int:
        ans=0
        for item in items:
            if ruleKey=='type' and item[0]==ruleValue:
                ans+=1
            elif ruleKey=='color' and item[1]==ruleValue:
                ans+=1
            elif ruleKey=='name' and item[2]==ruleValue:
                ans+=1
            else:
                continue
        return ans

java代码

class Solution {
    public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
        String x[]={"type","color","name"};
        int count=0,n=0;
        for(int i=0;i<3;i++){
            if(ruleKey.equals(x[i])){
                n=i;
                break;
            }
        }
        for(int i=0;i<items.size();i++){
            if(items.get(i).get(n).equals(ruleValue)){
                count=count+1;
            } 
        }
        return count;
    }
}

时间为4ms

C语言代码

int countMatches(char *** items, int itemsSize, int* itemsColSize, char * ruleKey, char * ruleValue){
    int i,n,ans=0;
    if(strcmp(ruleKey,"type")==0){
        n=0;
    }
    else if(strcmp(ruleKey,"color")==0){
        n=1;
    }
    else{
        n=2;
    }
    for(i=0;i<itemsSize;i++){
        if(strcmp(items[i][n],ruleValue)==0){
                ans++;
        }
    }
    return ans;
}

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2022-11-03，如有侵权请联系 cloudcommunity@tencent.com 删除

python