如何在Ballerina中求两个数组的交集并创建allrounders数组？

要找出同时属于batsmen和bowlers的球员（也就是allrounders），可以根据数组规模选择以下两种最优实现方式：

方法1：filter + exists（适合小规模数组） #

这种方式代码简洁直观，直接遍历batsmen数组，检查每个球员的id是否存在于bowlers数组中：

public type Player record {
    int id;
    string name;
};

Player[] batsmen = [
    {id: 1, name: "Virat"},
    {id: 2, name: "Rohit"},
    {id: 3, name: "Jadeja"}
];

Player[] bowlers = [
    {id: 3, name: "Jadeja"},
    {id: 4, name: "Bumrah"},
    {id: 5, name: "Shami"}
];

Player[] allrounders = batsmen.filter(b => bowlers.exists(bo => bo.id == b.id));

方法2：哈希集合优化（适合大规模数组） #

如果数组元素较多，用哈希集合存储bowlers的id可以将查找时间从O(n)降到O(1)，整体性能从O(n*m)提升到O(n+m)：

public type Player record {
    int id;
    string name;
};

Player[] batsmen = [
    {id: 1, name: "Virat"},
    {id: 2, name: "Rohit"},
    {id: 3, name: "Jadeja"}
];

Player[] bowlers = [
    {id: 3, name: "Jadeja"},
    {id: 4, name: "Bumrah"},
    {id: 5, name: "Shami"}
];

// 先提取bowlers的id存入哈希集合
hashset<int> bowlerIdSet = new(bowlers.map(bo => bo.id));
// 筛选出id在集合中的batsmen
Player[] allrounders = batsmen.filter(b => bowlerIdSet.has(b.id));

注意事项 #

优先用id作为判断依据，而非直接比较整个Player record：

• id是唯一标识，避免因名字重复导致的错误匹配
• 对比单个int字段比对比整个record的所有字段更高效

内容的提问来源于stack exchange，提问作者LUHEERATHAN THEVAKUMAR