動態遍歷列表

Question

我有一個如下所示的方法：

List<AbcObject> aaList= session.createCriteria(AbcObject.class)

aaList包含以下元素：

t_id   value
11      3
12      20
14      60
15      27    ------->(3+20+60+27=100), here min=11 & max=15
18      40  
22      20
33      40
45      20   -------->(40+20+40+20=100), here min=18 & max=45

我必須對列表進行迭代，以便如果value的運行總和達到100，則將一個元素添加到另一個列表中，如下所示：

min     max
11      15
18      45

請注意，完成除法運算的100值必須是可配置的。

也有可能第一個元素本身的值為100。在這種情況下， min和max的值相同。

我想出了以下方法：

final int BARRIER = 100;

//I am going to assume there are getter methods in your AbcObject and 
//that all values are greater than 0
List<int[]> minMaxs = new List<int[]>();
int sum = 0; 
int min = 0;
for (AbcObject obj: aaList) {
    if (sum == 0) {
        //start over, remember min
        min = obj.getT_id();
    }
    //add value to sum
    sum += obj.getValue();
    if (sum >= BARRIER) {
        //now we need to start again, remember what we got
        minMaxs.add(new int[]{min, obj.getT_id()});
        //reset
        min = 0;
        sum = 0;
    }
}

我發現的另一種方法是：

final int BARRIER = 100;

//I am going to assume there are getter methods in your AbcObject and 
//that all values are greater than 0
List<AbcObject> minMaxs = new List<AbcObject>();
int sum = 0; 
int min = 0;
for (AbcObject obj: aaList) {
    if (sum == 0) {
        //start over, remember min
        min = obj.getT_id();
    }
    //add value to sum
    sum += obj.getValue();
    if (sum >= BARRIER) {
        //now we need to start again, remember what we got
        minMaxs.add(new AbcObject(min, obj.getT_id()));
        //reset
        min = 0;
        sum = 0;
    }
}

有沒有更好的方法？

Answer 1

您的代碼看起來非常合理，而且易於理解。 我認為沒有迫切需要進行更改。