使用Java检查十进制数字顺序的逻辑

Question

I want to check the sequential order of decimal numbers and find the missing number. For eg: If i have 1.1.1, 1.1.3, 1.1.4, 2.1.1, 2.1.3, 2.1.2, 3, etc Here i need to find the missing number 1.1.2 and also out of sequence 2.1.2. Kindly help me with logic.

Answer 1

This does sound suspiciously like homework but, here's some hints for the algorithm. For simplicity, not efficiency, try a 2-step approach.

You'll have to treat each value in your initial list as an ordered set of integers. That is the value 2.1.3 is, an ArrayList whose elements are 2, 1, 3.

First determine what's out of sequence - this catches the 2.1.2 value. Something's out of sequence when the value of any part of the n-th element of the list is greater than any part of the (n+1)-th element. Walk through the list of values comparing two at a time; breaking each element into an list of integers.

Second, sort the list and determine if there are gaps. Sorting would still need to treat each value as a set of integers. A gap in the sorted list would be defined as a change of more than 1 in any part of the two values. Stop comparing 2 values when you find a gap and move onto the next 2 values to compare.