from和to

[bibi7]

//输入
const array = [
  {from:4, to:5}, 
  {from:3, to:4}, 
  {from:10, to:6}, 
  {from:5, to:10}, 
  {from:6, to:7}
]
//输出
//[{from:3, to:4}, {from:4, to:5}, {from:5, to:10}, {from:10, to:6}, {from:6, to:7}]

一开始想的是先找出最小的from，然后无脑用from去硬匹配to就行，不过想了想总感觉不大对，比如有下面这种情况：

//输入
const array = [
  {from:4, to:5}, 
  {from:18, to:4}, 
  {from:10, to:6}, 
  {from:5, to:10}, 
  {from:6, to:7}
]
//输出
//[{from:18, to:4}, {from:4, to:5}, {from:5, to:10}, {from:10, to:6}, {from:6, to:7}]

笨比了，根本不是要找最小的from，更类似于链表的形态，不存在to的from才是起点。还找个鸡儿最小from。

笨比解法开始：

function sortArray(array) {
  const fromArray = [];
  const toArray = [];
  for (let i = 0; i < array.length; i++) {
    const item = array[i];
    fromArray.push(item.from)
    toArray.push(item.to)
  }
  const resultFirstIndex= fromArray.find(item => toArray.indexOf(item) === -1)
  const resultFirstObj = array.find(item => item.from === resultFirstIndex)
  const result = []
  result.push(resultFirstObj)
  while (result.length < array.length) {
    result.push(array.find(item => item.from === result[result.length - 1]['to']))
  }
  return result
}

验证下：