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 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
    }
}
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