判断时间或者数字是否连续

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/dc282614966/article/details/85698741

一、判断时间数组是否连续（格式：yyyyMMdd）

public static String convert(String[] ints, int index) {
    int end = index;
    if (ints.length == index) {//结束条件，遍历完数组
        return "";
    } else {
        for (int i = index; i < ints.length; i++) {
            if (i < ints.length - 1) {
                Date date1 = DateUtils.addDate(DateUtils.format(String.valueOf(ints[i]),"yyyyMMdd"),1);
                Date date2 = DateUtils.format(String.valueOf(ints[i + 1]),"yyyyMMdd");
                if (date1.getTime()==date2.getTime()){
                    end = i;
                }
                else {
                    if (i > index)
                        end = end + 1;
                    break;
                }
            } else {
                if (end == ints.length - 2) {
                    end = ints.length - 1;
                    break;
                }
            }
        }
        if (index == end)//相等说明不连续
            return ints[index] + "," + convert(ints, end + 1);
        else//连续
            return ints[index] + "-" + ints[end] + "," + convert(ints, end + 1);
    }
}

二、判断数字数组是否连续

 public static String convert(Integer[] ints, int index) {
        int end = index;
        if (ints.length == index) {//结束条件，遍历完数组
            return "";
        } else {
            for (int i = index; i < ints.length; i++) {
                if (i < ints.length - 1) {
                    if (ints[i] + 1 == ints[i + 1]) {
                        end = i;
                    } else {
                        if (i > index)
                            end = end + 1;
                        break;
                    }
                } else {
                    if (end == ints.length - 2) {
                        end = ints.length - 1;
                        break;
                    }
                }
            }
            if (index == end)//相等说明不连续
                return ints[index] + "," + convert(ints, end + 1);
            else//连续
                return ints[index] + "-" + ints[end] + "," + convert(ints, end + 1);
        }
    }