FED47. 查找重复元素
描述
找出数组 arr 中重复出现过的元素(不用考虑返回顺序)示例1
输入:
[1, 2, 4, 4, 3, 3, 1, 5, 3]
输出:
[1, 3, 4]
HTML/CSS/JavaScript 解法, 执行用时: 841ms, 内存消耗: 77820KB, 提交时间: 2020-12-16
{"css":"","js":"function duplicates(arr) {\n var a = [],b = [];\n //遍历arr,如果以arr中元素为下标的的b元素已存在,则该b元素加1,否则设置为1\n for(var i = 0; i < arr.length; i++){\n if(!b[arr[i]]){\n b[arr[i]] = 1;\n continue;\n }\n b[arr[i]]++;\n }\n //遍历b数组,将其中元素值大于1的元素下标存入a数组中\n for(var i = 0; i < b.length; i++){\n if(b[i] > 1){\n a.push(i);\n }\n }\n return a;\n}","html":"","libs":[]}
HTML/CSS/JavaScript 解法, 执行用时: 842ms, 内存消耗: 77752KB, 提交时间: 2020-12-05
{"css":"","js":"function duplicates(arr) {\n var a=arr.sort(),b=[];\n for(var i in a){\n if(a[i]==a[i-1] && b.indexOf(a[i])==-1) b.push(a[i]); \n }\n return b;\n}","html":"","libs":[]}
HTML/CSS/JavaScript 解法, 执行用时: 843ms, 内存消耗: 77772KB, 提交时间: 2020-12-28
{"css":"","js":"function duplicates(arr) {\n var arr1 = [];\n arr.forEach(function (ele) {\n if (arr.indexOf(ele) != arr.lastIndexOf(ele) && arr1.indexOf(ele) == -1){\n arr1.push(ele);\n }\n });\n return arr1;\n}","html":"","libs":[]}
HTML/CSS/JavaScript 解法, 执行用时: 844ms, 内存消耗: 77772KB, 提交时间: 2021-03-28
{"css":"","js":" function duplicates(arr) {\n var temp = [];\n arr.forEach(function(elem){\n if(arr.indexOf(elem) != arr.lastIndexOf(elem) && temp.indexOf(elem) == -1){\n temp.push(elem);\n }\n });\n \n \n return temp;\n}","html":"","libs":[]}
HTML/CSS/JavaScript 解法, 执行用时: 845ms, 内存消耗: 77752KB, 提交时间: 2021-02-26
{"css":"","js":"function duplicates(arr) {\n var res = [];\n var hashmap = new Map();\n for(var num of arr){\n if(hashmap.has(num) === false) hashmap.set(num ,0);\n hashmap.set(num, hashmap.get(num)+1);\n }\n console.log(hashmap);\n for(var [key, val] of hashmap){\n if(val > 1) res.push(key);\n }\n return res;\n}","html":"","libs":[]}