Flutter：比較int和list <map<string,int> > </map<string,int>

Question

如果有人能夠提供幫助，謝謝。

這是我要與之比較的列表

[{"quote":3,"money":3000}, {"quote":6,"money":7000}, {"quote":8,"money":8000}]

在上一頁中，用戶輸入購買金額，例如 3500 美元。 如果數量超過

3000 then the user needs to present 3 quotes

7000 then the user needs to present 6 quotes

8000 then the user needs to present 8 quotes

我正在努力尋找一種解決方案來比較金額與列表值以獲得相應的 Map 值

如果有人有任何建議。 將不勝感激

謝謝

Answer 1

我將數據更改為更方便的結構

下面的所有代碼都將基於這個結構

final data = [{"quote":3,"money":3000}, {"quote":6,"money":7000}, {"quote":8,"money":8000},];

下一步是對數據進行排序，

如果數據已經像您提供的示例一樣排序，則可以跳過此步驟

我要使用合並排序對數據進行排序

我更改了Tomic-Riedel 的 mergeSort 實現以滿足我們的需要。

List<Map<String, int>> mergeSort(List<Map<String, int>> array) {

  // Stop recursion if array contains only one element
  if(array.length <= 1) {
    return array;
  }

  // split in the middle of the array
  int splitIndex = array.length ~/ 2;

  // recursively call merge sort on left and right array
  List<Map<String, int>> leftArray = mergeSort(array.sublist(0, splitIndex));
  List<Map<String, int>> rightArray = mergeSort(array.sublist(splitIndex));

  return merge(leftArray, rightArray);
}


List<Map<String, int>> merge(leftArray, rightArray) {
  List<Map<String, int>> result = [];
  int? i = 0;
  int? j = 0;

  // Search for the smallest eleent in left and right arrays
  // array and insert it into result
  // After the loop only one array has remaining elements
  while(i! < leftArray.length && j! < rightArray.length) {
    if(leftArray[i]['money'] <= rightArray[j]['money']) {
      result.add(leftArray[i]);
      i++;
    } else {
      result.add(rightArray[j]);
      j++;
    }
  }

  // Insert remaining elements of left array into result
  // as long as there are remaining elements
  while(i! < leftArray.length) {
    result.add(leftArray[i]);
    i++;
  }

  // Insert remaining elements of right array into result
  // as long as there are remaining elements
  while(j! < rightArray.length) {
    result.add(rightArray[j]);
    j++;
  }

  return result;
}

然后我們用它來獲取采購清單代碼：

final amount = 3500;
final sortedData = mergeSort(data);
// at this point out data is sorted by 'money'

List<Map<String, int>> canBePurchased = [];

for (Map<String, int> item in sortedData) {
  if (item['money']! <= amount) {
    canBePurchased.add(item);
  }
}
print(canBePurchased);

// to get only the last item
if (canBePurchased.isNotEmpty) {
  print(canBePurchased.last);
}