我需要先访问然后比较两个单独列表中的元素,并根据具有较小值的元素,删除较小元素并在第三个列表中为其分配位置。我该如何做到这一点?
答案 0 :(得分:1)
我相信您只需将列表中的元素添加到其中一个列表的新列表中,然后按升序对其进行排序即可。下面(伪代码):
List<Integer> list1 = Arrays.asList(2,4,5);
List<Integer> list2 = Arrays.asList(2,6,8,9,12,56);
List<Integer> mergedList = new ArrayList<Integer>();
mergedList.addAll(list1);
mergedList.addAll(list2);
Collections.sort(mergedList);
现在mergedList
应该有已排序的元素。
答案 1 :(得分:0)
仅供您学习,请在下方找到示例手动合并计划:
List<Integer> list1 = Arrays.asList(2,4,5);
List<Integer> list2 = Arrays.asList(2,6,8,9,12,56);
List<Integer> mergedList1 = new ArrayList<Integer>();
mergedList1.add(list1.get(0));
for(int indx = 1; indx< list1.size(); indx++){
int valueToMerge = list1.get(indx);
boolean merged = false;
for(int indx1 = 0; indx1< mergedList1.size(); indx1++){
if(mergedList1.get(indx1) > valueToMerge){
//add the value in middle
mergedList1.add(indx1, valueToMerge);
merged = true;
break;
}
}
if(!merged){
//add the value in the end
mergedList1.add(valueToMerge);
}
}
for(int valueToMerge: list2){
boolean merged = false;
for(int indx1 = 0; indx1< mergedList1.size(); indx1++){
if(mergedList1.get(indx1) > valueToMerge){
//add the value in middle
mergedList1.add(indx1, valueToMerge);
merged = true;
break;
}
}
if(!merged){
//add the value in the end
mergedList1.add(valueToMerge);
}
}