[LeetCode] Next Permutation - Next bigger number with the same set of digits

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replacement must be in-place, do not allocate extra memory. Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.

1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

思路:

You can do it in O(n) (where n is the number of digits) like this:

Starting from the right, you find the first pair-of-digits such that the left-digit is smaller than the right-digit. Let's refer to the left-digit by "digit-x". Find the smallest number larger than digit-x to the right of digit-x, and place it immediately left of digit-x. Finally, sort the remaining digits in ascending order - since they were already indescending order, all you need to do is reverse them (save for digit-x, which can be placed in the correct place in O(n)).

An example will make this more clear:

123456784987654321
start with a number

123456784 987654321
         ^the first place where the left-digit is less-than the right-digit is here.  
         Digit "x" is 4

123456784 987654321
              ^find the smallest digit larger than 4 to the right

123456785 4 98764321
        ^place it to the left of 4

123456785 4 12346789
123456785123446789
         ^sort the digits to the right of 5.  Since all of them except 
         the '4' were already in descending order, all we need to do is 
         reverse their order, and find the correct place for the '4'