二维数组中找到出现多次的元素

Question

var arr=new Array(["A","B"],["C","D"],["E","F"],["A","C"],["D","F"],["E","A"],["A","F"]);

var output_arr=new Array();

//找出出现过三次及以上的元素,写入到output_arr这个一维数组中

//"A"出现了4次，"F"出现了3次

output_arr = ["A","F"];

Answer 1

给你个参考- -let arr = new Array(["A", "B"], ["C", "D"], ["E", "F"], ["A", "C"], ["D", "F"], ["E", "A"], ["A", "F"]);let result = [],list = {},output_arr = [];计算元素出现次数并筛选出符合条件的元素  for (let [val1, val2] of arr) {    result.push(val1, val2);  }  for (let val of result) {    list[val] ? list[val] += 1 : list[val] = 1;    if (list[val] === 3) { //出现3次以上的元素      output_arr.push(val); //装进数组    }  }查看结果  console.log(output_arr);