Javascript 如何在一个对象Java脚本中获取不同键的重复值

function findDuplicatesSimple(obj) {
  let duplicates = {};
  let values = Object.values(obj);
  Object.keys(obj).forEach((key) => {
    // Check if the value appears more than once and hasn't been added to duplicates yet
    if (
      values.filter((v) => v.toLowerCase() === obj[key].toLowerCase()).length >
      1
    ) {
      duplicates[key] = obj[key];
    }
  });
  return duplicates;
}

const numn = {
  28: "The invincible special forces who returned to the Three Kingdoms(回到三国的无敌特种兵)",
  46: "Three Kingdoms: The opening incorporates Li Cunxiao(三国:开局融合了李存孝)",
  62: "Douluo: Super Beast Armed Guard Zhu Zhuqing(斗罗:超兽武装守护朱竹清)",
  76: "The Oriental Cavalry that swept across the Three Kingdoms(横扫三国的东方铁骑)",
  1514: "The Rise of an Empire(帝国崛起)",
  3140: "A leisurely life in another world(异界的悠闲生活)",
  5117: "The rise of an empire(帝国崛起)",
  5127: "After eavesdropping on my voice, the whole family went crazy(偷听我心声后，全家都杀疯了)",
  5148: "The Rise of an Empire(帝国崛起)",
  8440: "A ghost calls at the door in the middle of the night(夜半鬼叫门)",
  13140: "A leisurely life in another world(异界的悠闲生活)",
  13154: "The time travel story of Iron Ambition Man(钢铁雄心之舰男穿越记)",
};

console.log(findDuplicatesSimple(numn));

此函数迭代原始对象的键，并直接筛选这些值，以判断当前值是否在所有值中存在多次(不区分大小写).如果找到重复项，并且尚未将其添加到duplicates对象中，则会将其添加.这种略微简化的方法减少了代码量，但保留了有效解决问题的核心逻辑.

结果是:

{
  '1514': 'The Rise of an Empire(帝国崛起)',
  '3140': 'A leisurely life in another world(异界的悠闲生活)',
  '5117': 'The rise of an empire(帝国崛起)',
  '5148': 'The Rise of an Empire(帝国崛起)',
  '13140': 'A leisurely life in another world(异界的悠闲生活)'
}