java - 计算数组中连续项的出现

假设items数组由以下项目组成{3.1, 3.1, 3.1, 3.2, 3.2, 3.3, 3.4, 3.4, 3.4, 3.4, 3.1, 3.1}

我想要的是计算连续项目中每个项目的出现，例如：

3.1 = 3
3.2 = 2
3.3 = 1
3.4 = 4
3.1 = 2

我写了以下函数：

private void displayItems(List<Double> items) {
        double current_item=0;
        for(int i=0; i<items.size(); i++) {
            int count=1;
            current_item = items.get(i);
            if(i != items.size()) {
                for(int j=i+1; j<items.size(); j++) {
                    double next_item = items.get(j);
                    if(current_item == next_item) {
                        count++;
                    }else {
                        break;
                    }
                }
                System.out.println("item value is " + current_item + " and count is " + count);
            }
        }
    }

我得到以下结果：

item value is 3.1 and count is 3
item value is 3.1 and count is 2
item value is 3.1 and count is 1
item value is 3.2 and count is 2
item value is 3.2 and count is 1
item value is 3.3 and count is 1
item value is 3.4 and count is 4
item value is 3.4 and count is 3
item value is 3.4 and count is 2
item value is 3.4 and count is 1
item value is 3.1 and count is 2
item value is 3.1 and count is 1

我该怎么做才能显示如下结果：

item value is 3.1 and count is 3
item value is 3.2 and count is 2
item value is 3.3 and count is 1
item value is 3.4 and count is 4
item value is 3.1 and count is 2

请不要说我不想统计整个数组中每个项目的出现，我只想统计其在连续项目中的出现。

最佳答案

您的代码正在迭代先前迭代中已经计算过的值。逻辑上的小调整可以按预期进行。

private void displayItems(List<Double> items) {
        double current_item=0;
        for(int i=0; i<items.size(); i++) {
            int count=1;
            current_item = items.get(i);
            if(i != items.size()) {
                int j=i+1;
                for(; j<items.size(); j++) {
                    double next_item = items.get(j);
                    if(current_item == next_item) {
                        count++;
                    }else {
                        break;
                    }
                }
                System.out.println("item value is " + current_item + " and count is " + count);
                i = j-1;
            }
        }
    }

关于java - 计算数组中连续项的出现，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/49635017/