JavaScript—改进动态对象的构造

我有一个构建对象的脚本：

console.log(solution([2, 1, 3, 5, 3, 2]))


function solution(inputArray) {
  let arrMap = [] 

  for (let i = 0; i < inputArray.length; i++) { 
      var repeated = false;
      var e = arrMap.find(s => s.element == inputArray[i])
      
      if(e){        
        repeated = true;
        e.repeated = true;      
        e.firstPosition = e.position
        e.lastPosition = i           
        delete e['position'];
      }
      
      arrMap.push({
        element : inputArray[i],
        position: i,       
        repeated: repeated
      })
    } 

  return arrMap
}

我只是想通过删除重复的项目的剩余行（在本例中为项目2和3）来更改报税表。如下图所示：

[
   { element: 2, repeated: true, firstPosition: 0, lastPosition: 5 },
   { element: 1, position: 1, repeated: false },
   { element: 3, repeated: true, firstPosition: 2, lastPosition: 4 },
   { element: 5, position: 3, repeated: false },
   { element: 3, position: 4, repeated: true }, //<---- DUPLICATED! DELETE THIS LINE
   { element: 2, position: 5, repeated: true } //<---- DUPLICATED! DELETE THIS LINE
]

 [
   { element: 2, repeated: true, firstPosition: 0, lastPosition: 5 },
   { element: 1, position: 1, repeated: false },
   { element: 3, repeated: true, firstPosition: 2, lastPosition: 4 },
   { element: 5, position: 3, repeated: false },     
 ]

---

减少到中间对象使检查元素是否已添加到结果中变得更容易（而且更便宜，性能更高）：

const solution=（a）=>Object.values（a.reduce（（a，v，i）=>{
if（a[v]）{
a[v]。重复=真；
a[v].firstPosition=a[v].firstPosition | | a[v].位置；
a[v].lastPosition=i；
删除[v]位置；
}否则{
a[v]={
要素：v，
重复：错，
职位：i
};
}
返回a；
}, {}));
log（解决方案（[2,1,3,5,3,2]）我根据用户120242的想法改进了代码，现在它运行良好，达到了预期的效果

function solution(inputArray) {
  let arrMap = []

  for (let i = 0; i < inputArray.length; i++) {   
    var e = arrMap.find(s => s.element == inputArray[i])

    if (e) {
      e.repeated = true;
      e.firstPosition = e.position
      e.lastPosition = i
      delete e['position'];
    } else {
      arrMap.push({
        repeated: false,
        element: inputArray[i],
        position: i,
        repeated: repeated
      })
    }
  }

  return arrMap
}

函数解决方案（输入阵列）{
让arrMap=[]
对于（设i=0；is.element==inputArray[i]）
如果（e）{
e、 重复=正确；
e、 firstPosition=e.position
e、 lastPosition=i
删除e[‘位置’]；
}否则{
arrMap.push({
重复：错，
元素：输入阵列[i]，
职位：我，
重复：重复
})
}
}
返回arrMap
}
只需在
arrMap.push之前添加一个
else
，这样只有在没有找到重复项时才会进行推送。您的代码当前总是尝试添加新元素。
function solution(inputArray) {
  let arrMap = []

  for (let i = 0; i < inputArray.length; i++) {   
    var e = arrMap.find(s => s.element == inputArray[i])

    if (e) {
      e.repeated = true;
      e.firstPosition = e.position
      e.lastPosition = i
      delete e['position'];
    } else {
      arrMap.push({
        repeated: false,
        element: inputArray[i],
        position: i,
        repeated: repeated
      })
    }
  }

  return arrMap
}