循环直到在 Javascript 中找到一个空数组

Question

我正在尝试找到一个解决方案来进行循环 (Javascript)，直到对象数组为空。这是我要使用的对象：

"chain": {
    "evolves_to": [{
        "evolves_to": [{
            "evolves_to": [],
            "species": {
                "name": "nidoqueen"
            }
        }],
        "species": {
            "name": "nidorina"
        }
    }],
    "species": {
        "name": "nidoran-f"
    }
}

我想循环直到发现变量 evolves_to 为空。并在每个循环中使用 species.name 来列出进化，在我的例子中： nidoran-f -> nidorina -> nidoqueen

我还没有找到好的方法。有点失落。感谢您的帮助 ;)

Answer 1

你可以使用递归函数：

const chain = {
    "evolves_to": [{
        "evolves_to": [{
            "evolves_to": [],
            "species": {
                "name": "nidoqueen"
            }
        }],
        "species": {
            "name": "nidorina"
        }
    }],
    "species": {
        "name": "nidoran-f"
    }
}
, traverse = obj => {
  if(!obj.evolves_to.length) {
    console.log(obj.species.name)
    return
  } else {
    console.log(obj.species.name, "=>")
    traverse(obj.evolves_to[0])    
  }
}

traverse(chain)

或者在数组中收集值：

const chain = {
    "evolves_to": [{
        "evolves_to": [{
            "evolves_to": [],
            "species": {
                "name": "nidoqueen"
            }
        }],
        "species": {
            "name": "nidorina"
        }
    }],
    "species": {
        "name": "nidoran-f"
    }
}
, arr = []
, traverse = obj => {
  if(!obj.evolves_to.length) {
    arr.push(obj.species.name)
    return
  } else {
    arr.push(obj.species.name)
    traverse(obj.evolves_to[0])    
  }
}

traverse(chain)
console.log(arr.join(" -> "))

Answer 2

一个简单的递归函数应该可以解决问题：

const evolutions = { 
  "chain": {
    "evolves_to": [{
        "evolves_to": [{
            "evolves_to": [],
            "species": {
                "name": "nidoqueen"
            }
        }],
        "species": {
            "name": "nidorina"
        }
    }],
    "species": {
        "name": "nidoran-f"
    }
  }
}

function findNoEvolution(obj, evolutionStages) {
  if (obj.evolves_to.length > 0) {
    for (let i = 0; i < obj.evolves_to.length; i += 1) {
      const found = findNoEvolution(obj.evolves_to[i], evolutionStages);
      if (found) {
        evolutionStages.push(obj.species.name)
        return found;
      }
    }
  }
  evolutionStages.push(obj.species.name)
  return true;
}

const evolutionStages = [];
findNoEvolution(evolutions.chain, evolutionStages);
const evolutionStagesStr = evolutionStages.reverse().join(' -> ');
console.log(evolutionStagesStr);

然后将结果进化数组反转为通过reverse()从第一次进化开始，并通过joing(' -> ')用箭头连接。

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