In a deck of cards, each card has an integer written on it. Return`true`if and only if you can choose?`X >= 2`such that?it is possible to split the entire deck?into 1 or more groups of cards, where: * Each group has exactly`X`cards. * All the cards in each group have the same integer. **Example 1:** ~~~ Input: [1,2,3,4,4,3,2,1] Output: true Explanation: Possible partition [1,1],[2,2],[3,3],[4,4] ~~~ **Example 2:** ~~~ Input: [1,1,1,2,2,2,3,3] Output: false Explanation: No possible partition. ~~~ **Example 3:** ~~~ Input: [1] Output: false Explanation: No possible partition. ~~~ **Example 4:** ~~~ Input: [1,1] Output: true Explanation: Possible partition [1,1] ~~~ **Example 5:** ~~~ Input: [1,1,2,2,2,2] Output: true Explanation: Possible partition [1,1],[2,2],[2,2] ~~~ **Note:** 1. `1 <= deck.length <= 10000` 2. `0 <= deck[i] <?10000` ``` var hasGroupsSizeX = function(deck) { // 最大公約數計算公式 function gcd(num1, num2) { // 利用輾轉相除法來計算最大公約數 return num2 === 0 ? num1 : gcd(num2, num1 % num2); } // 相同牌出現次數Map let timeMap = new Map(); // 遍歷牌 deck.forEach(num => { // 統計每張牌出現的次數 timeMap.set(num, timeMap.has(num) ? timeMap.get(num) + 1 : 1); }); // Map.protype.values()返回的是一個新的Iterator對象， 所以可以使用擴展運算符(...)來構造成數組 let timeAry = [...timeMap.values()]; /* 最大公約數 因為該數組是出現次數數組，最小值至少為1（至少出現1次）， 所以默認賦值為數組首位對公約數計算無干擾 */ let g = timeAry[0]; // 遍歷出現次數，計算最大公約數 timeAry.forEach(time => { // 因為需要比較所有牌出現次數的最大公約數，故需要一個中間值 g = gcd(g, time); }); // 判斷是否滿足題意 return g >= 2; }; ```