在實際工作或面試中，我們經(jīng)常會遇到“數(shù)組去重”問題，接下來就是使用js實現(xiàn)的數(shù)組去重的多種方法：

1.將數(shù)組的每一個元素依次與其他元素做比較，發(fā)現(xiàn)重復(fù)元素，刪除

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5,5,5,5]; console.log(arr); //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5, 5, 5, 5] function noRepeat1(arr) {for(var i = 0; i < arr.length-1; i++){ for(var j = i+1; j < arr.length; j++){if(arr[i]===arr[j]){ arr.splice(j,1); j--;} }}return arr; } var arr2 = noRepeat1(arr); console.log(arr2); //[1, 23, 3, 5, 6, 7, 9, 8]

2.借助indexOf()方法判斷此元素在該數(shù)組中首次出現(xiàn)的位置下標(biāo)與循環(huán)的下標(biāo)是否相等

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5,5,5]; console.log(arr); //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5, 5, 5] function noRepeat2(arr) {for (var i = 0; i < arr.length; i++) { if (arr.indexOf(arr[i]) != i) {arr.splice(i,1);//刪除數(shù)組元素后數(shù)組長度減1后面的元素前移i--;//數(shù)組下標(biāo)回退 }}return arr; } var newArr = noRepeat2(arr); console.log(newArr); //[1, 23, 3, 5, 6, 7, 9, 8]

3.利用數(shù)組中的filter方法

var arr = [’apple’,’banana’,’pear’,’apple’,’orange’,’orange’];console.log(arr) //['apple', 'banana', 'pear', 'apple', 'orange', 'orange']var newArr = arr.filter(function(value,index,self){ return self.indexOf(value) === index;});console.log(newArr); //['apple', 'banana', 'pear', 'orange']

4.借助新數(shù)組 通過indexOf方判斷當(dāng)前元素在數(shù)組中的索引如果與循環(huán)的下標(biāo)相等則添加到新數(shù)組中

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5,5,5]; console.log(arr) //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5, 5, 5] function noRepeat4(arr) {var ret = [];for (var i = 0; i < arr.length; i++) { if (arr.indexOf(arr[i]) == i) {ret.push(arr[i]); }}return ret; } var arr2 = noRepeat4(arr); console.log(arr2); //[1, 23, 3, 5, 6, 7, 9, 8]

5.利用空對象來記錄新數(shù)組中已經(jīng)存儲過的元素

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5]; console.log(arr) //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5] var obj={}; var newArr=[]; for(var i=0;i<arr.length;i++){if(!obj[arr[i]]){ obj[arr[i]]=true; newArr.push(arr[i]);} } console.log(newArr); //[1, 23, 3, 5, 6, 7, 9, 8]

6.借助新數(shù)組，判斷新數(shù)組中是否存在該元素如果不存在則將此元素添加到新數(shù)組中

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5]; console.log(arr); //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5] function noRepeat6(arr){var newArr = [];for(var i = 0; i < arr.length; i++){ if(newArr.indexOf(arr[i]) == -1){newArr.push(arr[i]); }}return newArr; } var arr2 = noRepeat6(arr); console.log(arr2); //[1, 23, 3, 5, 6, 7, 9, 8]

7.借助新數(shù)組，判斷新數(shù)組中是否存在該元素如果不存在則將此元素添加到新數(shù)組中（原數(shù)組長度不變但被按字符串順序排序）

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5]; console.log(arr); //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5] function noRepeat7(arr) {var ret = [], end;//臨時變量用于對比重復(fù)元素arr.sort();//將數(shù)重新組排序end = arr[0];ret.push(arr[0]);for (var i = 1; i < arr.length; i++) { if (arr[i] != end) {//當(dāng)前元素如果和臨時元素不等則將此元素添加到新數(shù)組中ret.push(arr[i]);end = arr[i]; }}return ret; } var arr2 = noRepeat7(arr); console.log(arr2); //[1, 23, 3, 5, 6, 7, 8, 9]

8.此方法沒有借助新數(shù)組直接改變原數(shù)組,并且去重后的數(shù)組被排序

var arr = [1,23,1,1,1,3,23,5,6,7,9,9,8,5]; console.log(arr); //[1, 23, 1, 1, 1, 3, 23, 5, 6, 7, 9, 9, 8, 5] function noRepeat8(arr) {var end;//臨時變量用于對比重復(fù)元素arr.sort();//將數(shù)重新組排序end = arr[0];for (var i = 1; i < arr.length; i++) { if (arr[i] == end) {//當(dāng)前元素如果和臨時元素相等則將此元素從數(shù)組中刪除arr.splice(i,1);i--; }else{end = arr[i]; }}return arr; } var arr2 = noRepeat8(arr); console.log(arr2); //[1, 23, 3, 5, 6, 7, 8, 9]

9.雙層循環(huán)改變原數(shù)組

var arr = [1,1,2,2,3,3,4,4,5,5,4,3,1,2,6,6,6,6]; console.log(arr); //[1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 1, 2, 6, 6, 6, 6] function noRepeat9(arr){for (var i = 0; i < arr.length; i++) { for (var j = 0; j < arr.length; j++) {if (arr[i] == arr[j] && i != j) {//將后面重復(fù)的數(shù)刪掉 arr.splice(j, 1);} }}return arr; } var arr2 = noRepeat9(arr); console.log(arr2); //[1, 2, 3, 4, 5, 6]

10.借助新數(shù)組

var arr = [1,1,2,2,3,3,4,4,5,5,4,3,2,1,1,1]; console.log(arr); //[1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 2, 1, 1, 1] var newArr = []; for (var i = 0; i < arr.length; i++) {var repArr = [];//接收重復(fù)數(shù)據(jù)后面的下標(biāo)//內(nèi)層循環(huán)找出有重復(fù)數(shù)據(jù)的下標(biāo)for (var j = i + 1; j < arr.length; j++) { if (arr[i] == arr[j]) {repArr.push(j);//找出后面重復(fù)數(shù)據(jù)的下標(biāo) }}//console.log(repArr);if (repArr.length == 0) {//若重復(fù)數(shù)組沒有值說明其不是重復(fù)數(shù)據(jù) newArr.push(arr[i]);} } console.log(newArr); //[5, 4, 3, 2, 1]

11.借助ES6提供的Set結(jié)構(gòu)

var arr = [1,1,2,2,3,3,4,4,5,5,4,3,2,1,1,1]; console.log(arr); //[1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 2, 1, 1, 1] function noRepeat11(arr){var newArr = [];var myset = new Set(arr);//利用了Set結(jié)構(gòu)不能接收重復(fù)數(shù)據(jù)的特點for(var val of myset){ newArr.push(val)}return newArr; } var arr2 = noRepeat11(arr) console.log(arr2); //[1, 2, 3, 4, 5]

以上就是js數(shù)組去重的11種方法的詳細內(nèi)容，更多關(guān)于js數(shù)組去重的資料請關(guān)注好吧啦網(wǎng)其它相關(guān)文章！