java解决统计匹配规则的物品数量
01 题目
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"- 所有字符串仅由小写字母组成
02 知识点
- String类型的比较
- list有序集合
03 我的题解思路
public class countMatches {public static void main(String[] args) {
// 测试数据List<List<String>> items=Arrays.asList(
Arrays.asList ("phone","blue","pixel"),
Arrays.asList ("computer","silver","lenovo") ,
Arrays.asList ("phone","gold","iphone"));System.out.println(countMatches(items, "color", "red"));}public static int countMatches(List<List<String>> items, String ruleKey, String ruleValue) {int count=0;//记录符合条件的值int num=0;//判断ruleKey要求的索引下标if(ruleKey.equals("type")) {num=0;}else if (ruleKey.equals("color")) {num=1;}else if (ruleKey.equals("name")) {num=2;}for (int i = 0; i < items.size(); i++) {
// 从集合循环取出并判断对应索引if(items.get(i).get(num).equals(ruleValue)) {count++;}}return count;}
}
![[蓝桥杯2018初赛]耐摔指数 （动态规划）](http://pic.xiahunao.cn/[蓝桥杯2018初赛]耐摔指数 （动态规划）)
)
