// 最簡單數組去重法/** 新建一新數組，遍歷傳入數組，值不在新數組就push進該新數組中* IE8以下不支持數組的indexOf方法*/function uniq(array){  var temp = []; //一個新的臨時數組  for(var i = 0; i < array.length; i++){    if(temp.indexOf(array[i]) == -1){      temp.push(array[i]);    }  }  return temp;}var newArray = [1,2,2,4,9,6,7,5,2,3,5,6,5];console.log(uniq(newArray));// [1, 2, 4, 9, 6, 7, 5, 3]/*var newArray = ['mu','zi','dig','big','z','mu','zi','muzidigbig']console.log(uniq(newArray));//["mu", "zi", "dig", "big", "z", "muzidigbig"]*/

/** 還是得調用“indexOf”性能跟方法1差不多，* 實現思路：如果當前數組的第i項在當前數組中第一次出現的位置不是i，* 那么表示第i項是重復的，忽略掉。否則存入結果數組。*/function uniq(array){  var temp = [];  for(var i = 0; i < array.length; i++) {    //如果當前數組的第i項在當前數組中第一次出現的位置是i，才存入數組；否則代表是重復的    if(array.indexOf(array[i]) == i){      temp.push(array[i])    }  }  return temp;}var aa = [1,2,"2",4,9,"a","a",2,3,5,6,5];console.log(uniq(aa));//[1, 2, "2", 4, 9, "a", 3, 5, 6]

/** 給傳入數組排序，排序后相同值相鄰，* 然后遍歷時,新數組只加入不與前一值重復的值。* 會打亂原來數組的順序*/function uniq(array){  array.sort();  console.log(array.sort());  var temp=[array[0]];  for(var i = 1; i < array.length; i++){    if( array[i] !== temp[temp.length-1]){      temp.push(array[i]);    }  }  return temp;}var aa = [1,2,"2",4,9,"a","a",2,3,5,6,5];console.log(uniq(aa));//[1, "2", 2, 3, 4, 5, 6, 9, "a"]

// 思路：獲取沒重復的最右一值放入新數組/** 推薦的方法* 方法的實現代碼相當酷炫，* 實現思路：獲取沒重復的最右一值放入新數組。* （檢測到有重復值時終止當前循環同時進入頂層循環的下一輪判斷）*/function uniq(array){  var temp = [];  var index = [];  var l = array.length;  for(var i = 0; i < l; i++) {    for(var j = i + 1; j < l; j++){      if (array[i] === array[j]){        i++;        j = i;      }    }    temp.push(array[i]);    index.push(i);  }  console.log(index);//[0, 2, 5, 6, 7, 8, 9]  return temp;}var aa = [1,2,2,3,5,3,6,5,'3','muzi'];console.log(uniq(aa));// [1, 2, 3, 6, 5, "3", "muzi"]

//利用數組原型對象上的includes方法function unique(arr){ var bArr= []; for(var i=0; i<arr.length; i++){  if( !bArr.includes(arr[i]) ){ // 如果bArr新數組包含當前循環item   bArr.push(arr[i]);  } } return bArr;}var aa = [1,2,2,3,5,3,6,5,'3','muzi'];console.log(unique(aa))//[1, 2, 3, 5, 6, "3", "muzi"]

//利用數組原型對象上的filter和includes方法function unique6(arr){ var bArr= []; bArr= arr.filter(function(item){ return bArr.includes(item) ? null : bArr.push(item); }); return bArr;}var aa = [1,2,2,3,5,3,6,5,'3','muzi'];console.log(unique6(aa))//[1, 2, 3, 5, 6, "3", "muzi"]

//利用數組原型對象上的forEach和includes方法function unique7(arr){ var bArr= []; arr.forEach(function(item){  bArr.includes(item) ? null : bArr.push(item); });  return bArr;}var aa = [1,2,2,3,5,3,6,5,'3','muzi'];console.log(unique7(aa))//[1, 2, 3, 5, 6, "3", "muzi"]

  Array.prototype.method1 = function(){    var arr=[];  //定義一個臨時數組    for(var i = 0; i < this.length; i++){  //循環遍歷當前數組      //判斷當前數組下標為i的元素是否已經保存到臨時數組      //如果已保存，則跳過，否則將此元素保存到臨時數組中      if(arr.indexOf(this[i]) == -1){        arr.push(this[i]);      }    }    return arr;  };  var arrNum1 = [1,4,1,1,3,3,4,6,7,8,3,7,0,2,11,2,'muzi','digbig','muzi'];  console.log(arrNum1.method1());//[1, 4, 3, 6, 7, 8, 0, 2, 11, "muzi", "digbig"]

  let arr = [1,2,2,3,4,4,4];  let s = new Set(arr);  //結果：Set {1,2,3,4}  let newArr = Array.from(s);  //結果：[1,2,3,4],完成去重