我有一个嵌套的数组,一旦满足条件,它应该给所有的父代码,例如我有一个数据数组,我应该在其中匹配
> getParentIds(data,182,[]);
>结果：[96,182];
> getParentIds(data,174,[]);
>结果：[109,219,76,174];
var data = [{
"id": 96,"name": "test1","items": [{
"id": 181,"name": "Yes","items": []
},{
"id": 182,"name": "No","items": []
}]
},{
"id": 109,"name": "Test5","items": [{
"id": 219,"name": "opt2","items": [{
"id": 76,"name": "test3","items": [{
"id": 173,"items": []
},{
"id": 174,"items": [{
"id": 100,"name": "test2","items": [{
"id": 189,"items": []
}]
}]
}]
}]
},{
"id": 224,"name": "opt3","items": []
}]
}];
function getParentIds(data,id,parentIds) {
if (!parentIds) {
parentIds = [];
}
data.map(function(item) {
if (item.id === id) {
parentIds.push(item.id);
return parentIds;
} else if (item.items.length === 0) {
// do nothing
} else {
return getParentIds(item.items,parentIds);
}
});
}
console.log("Array list: " + getParentIds(data,[]));
你能给我任何建议吗？
解决方法
这是一个很酷的问题.我花了比我预期的更多,但这是一个
breadth-first search的实现：
var data = [{
"id": 96,"items": []
}]
}];
function parentsOf( arr,parents){
if (parents.length)
return parents;
// I use for(;;) instead of map() because I need the return to exit the loop
for (var i = 0; i < arr.length; i++){
if ( arr[i].id == id){
//push the current element at the front of the parents array
parents.unshift( arr[i].id );
return parents;
};
if ( arr[i].items ){
parents = parentsOf(arr[i].items,parents);
// if the parents array has any elements in it it means we found the child
if (parents.length){
parents.unshift(arr[i].id);
return parents;
}
}
}
return parents;
}
console.log("Array list for 182: " + parentsOf(data,[]));
console.log("Array list for 174: " + parentsOf(data,[]));