比较两个复杂的字符串数组列表[关闭]

Question

我有两个ArrayLists，其中存储了字符串值。 列表可以是：大小不同，每个列表具有相同的值和不同的不匹配字符串。

例子：

List 1: "a","b","b","c","e","a","e","f","g";

List 2: "a","a","b","c","a","e","a","d","f","e","e","g";

我想要得到的结果是列表 1 与列表 2 相比的匹配值。所以示例的结果是：

结果：

[String] [pos L1] [pos L2]

["a"] [0] [0]

["b"] [1] [2]

["c"] [3] [3]

["e"] [4] [5]

["a"] [5] [6]

["e"] [6] [9]

["g"] [8] [11]

Answer 1

执行两个嵌套循环并从您在内部循环的最后一次迭代中停止的索引开始在第二个循环中进行迭代：

public static List<String> algo(List<String> l1, List<String> l2) {
        List<String> result = new ArrayList<>();

        int lastIndexFound = 0;

        for(int i = 0; i < l1.size(); i++) {
            for(int j = lastIndexFound; j < l2.size(); j++) {
                String list1element = l1.get(i);
                String list2element = l2.get(j);
                if(list1element.equals(list2element)) {
                    result.add(list1element);
                    lastIndexFound = j + 1;
                    break;
                }
            }
        }

        return result;
    }

用于测试：

List<String> list = Lists.newArrayList("a","b","b","c","e","a","e","f","g");
List<String> list2 = Lists.newArrayList("a","a","b","c","a","e","a","d","f","e","e","g");

algo(list, list2).forEach(System.out::println);

这个输出：

a
b
c
e
a
e
g

Answer 2

ArrayList<yourElementType> resultsArray = new Arraylist<yourElementType>()

for (yourElementType e : List1){

if(List2.contains(e)){
resultsArray.add(e)
// if you want the exact position, use list2.indexof(e)
}

}