LeetCode（String)1773. Count Items Matching a Rule

落日映苍穹つ 2023-09-23 22:05 47阅读 0赞

# 1.问题 #

You are given an array items, where each items\[i\] = \[typei, colori, namei\] describes the type, color, and name of the ith item. You are also given a rule represented by two strings, ruleKey and ruleValue.

The ith item is said to match the rule if one of the following is true:

ruleKey == “type” and ruleValue == typei.  
ruleKey == “color” and ruleValue == colori.  
ruleKey == “name” and ruleValue == namei.  
Return the number of items that match the given rule.

Example 1:

> Input: items = \[\[“phone”,“blue”,“pixel”\],\[“computer”,“silver”,“lenovo”\],\[“phone”,“gold”,“iphone”\]\], ruleKey = “color”, ruleValue = “silver”  
> Output: 1  
> Explanation: There is only one item matching the given rule, which is \[“computer”,“silver”,“lenovo”\].

Example 2:

> Input: items = \[\[“phone”,“blue”,“pixel”\],\[“computer”,“silver”,“phone”\],\[“phone”,“gold”,“iphone”\]\], ruleKey = “type”, ruleValue = “phone”  
> Output: 2  
> Explanation: There are only two items matching the given rule, which are \[“phone”,“blue”,“pixel”\] and \[“phone”,“gold”,“iphone”\]. Note that the item \[“computer”,“silver”,“phone”\] does not match.

Constraints:

*  1 <= items.length <= 104
 *  1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
 *  ruleKey is equal to either “type”, “color”, or “name”.
 *  All strings consist only of lowercase letters.

# 2.解题思路 #

## 方法1： ##

> 1.定义返回值sum为0  
> 2.遍历items,如果ruleKey为type,color, name时，//ruleValue和取出\[0\]，\[1\]，\[2\]位置比较，两值相等+1  
> 3.返回sum的值

## 方法2： ##

> 1.定义index和sum的初始值为0  
> 2.使用 switch case 识别列表中的哪个索引，为了匹配ruleValue  
> 3 List item，结果显示一行数据，在根据索引的值比较是否相等，如果有相等元素+1 例：\[“phone”,“blue”,“pixel”\]  
> 4.返回sum的值

# 3.代码 #

## 代码1： ##

class Solution {
        
        public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
        
            int sum=0;//1.定义返回值sum为0
            for(int i =0;i<items.size();i++){
        //2.遍历items,如果ruleKey为type,color, name时，//ruleValue和取出[0]，[1]，[2]位置比较，两值相等+1
                if(ruleKey.equals("type")) {
         
                    if(ruleValue.equals(items.get(i).get(0))) sum++;
                }else if(ruleKey.equals("color")){
        
                    if(ruleValue.equals(items.get(i).get(1))) sum++;
                }else{
        
                    if(ruleValue.equals(items.get(i).get(2))) sum++;
                }
            }
            return sum;//3.返回sum的值
        }
    }

## 代码2： ##

public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
        
      int index = 0, sum = 0;//1.定义index和sum的初始值为0
            //2.使用 switch case 识别列表中的哪个索引，为了匹配ruleValue
            switch (ruleKey) {
        
                case "type":
                    index = 0;
                    break;
                case "color":
                    index = 1;
                    break;
                default:
                    index = 2;
            }
            for (List<String> item : items) {
        //3 List<String> item，结果显示一行数据，在根据索引的值比较是否相等，如果有相等元素+1    例：["phone","blue","pixel"]
                if (item.get(index).equals(ruleValue))
                    sum ++;
            }
            return sum;//4.返回sum的值
        }

> 和代码2解题思路相同，switch换成if进行判断

class Solution {
        
        public int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
        
            
          	int result = 0;
            int index = 0;
    
            if(ruleKey.equals("color")) index = 1;
            if(ruleKey.equals("name")) index = 2;
    
            for(int i  = 0 ; i< items.size() ; i++)
                if((items.get(i).get(index)).equals(ruleValue) ) result++;
            return result;
        }
    }

# IDEA本地运行代码 #

> 为了了解 List<List> items的数据结构，添加了运行在本地代码，方便学习和理解！

public class Test {
        
    
        public static void main(String args[]){
        
            List<List<String>> items = new ArrayList<>();
            List<String> l1 =new ArrayList<>();
            l1.add("phone");
            l1.add("blue");
            l1.add("pixel");
            List<String> l2 =new ArrayList<>();
            l2.add("computer");
            l2.add("silver");
            l2.add("lenovo");
            List<String> l3 =new ArrayList<>();
            l3.add("phone");
            l3.add("gold");
            l3.add("iphone");
    
            items.add(l1);
            items.add(l2);
            items.add(l3);
    
    
            String ruleKey = "color";
            String ruleValue ="silver";
            
            System.out.println(countMatches(items, ruleKey,ruleValue));
    
        }
    
        public static int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {
        
            int sum=0;//1.定义返回值sum为0
            for(int i =0;i<items.size();i++){
        //2.遍历items,如果ruleKey为type,color, name时，//ruleValue和取出[0]，[1]，[2]位置比较，两值相等+1
                if(ruleKey.equals("type")) {
        
                    if(ruleValue.equals(items.get(i).get(0))) sum++;
                }else if(ruleKey.equals("color")){
        
                    if(ruleValue.equals(items.get(i).get(1))) sum++;
                }else{
        
                    if(ruleValue.equals(items.get(i).get(2))) sum++;
                }
            }
            return sum;//3.返回sum的值
        }
    
    }