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This article mainly shows you how to use java to solve the next arrangement problem, the content is easy to understand, clear, hope to help you solve doubts, the following let the editor to take you to study and learn "how to use java to solve the next arrangement problem" this article.

To implement the function to get the next arrangement, the algorithm needs to rearrange the given sequence of numbers into the next larger arrangement in the lexicographic order.

If there is no next larger arrangement, the numbers are rearranged into the smallest arrangement (that is, ascending order).

It must be modified in place, and only extra constant space is allowed.

Here are some examples where the input is in the left column and the corresponding output is in the right column.

1,2,3 → 1,3,2

3,2,1 → 1,2,3

1,1,5 → 1,5,1

Parsing:

Dict is the number of occurrences of each word. Look for it in the while loop below. If the number of lookups for a word is greater than the number of occurrences of a word in words, the current loop will be terminated. The length of each word in words is the same, wordlen is the length of a single word, and totalLen is the length of all words in words. In the following for and while loops, it is equivalent to truncating a substring of length totalLen from the source string s, and then truncating a string of length wordLen from the substring and storing it in found each time, and terminating the while loop when a string in found is larger than a string in dict Otherwise, continue to search until the length of the search is greater than totalLen, then terminate the while loop, and then compare. If you still don't understand, you can also write this in the for loop, intercepting a string of length totalLen each time, and then intercepting a string of length wordLen with a string of length totalLen, which will most likely be compared with the string in words one by one.

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