[Leetcode] [JavaScript] Next permutation

Next permutation

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If Such arrangement is not a possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must is in-place, do not allocate extra memory.

Here is some examples. Inputs is in the left-hand column and its corresponding outputs is in the right-hand column.
1,2,3→1,3,2
3,2,1→1,2,3
1,1,5→1,5,1

https://leetcode.com/problems/next-permutation/

Find the next arrangement.

The size of the string is a bit to compare, the next arrangement is just larger than the current arrangement.

The simplest case is [1, 2, 3], and only the last two bits are exchanged to get the next sequence.

The complex case [1, 2, 4, 3], found that the last substring [4, 3] is already the largest, then need to move a number 2 to the front, and the back substring to be incremented [3, 2], the result is [4, 1, 3, 2].

Again such as [1, 4, 7, 5, 3, 2], the result is [1, 5, 2, 3, 4, 7]

The realization of the time, first judge is not descending sequence, if it is reverse all,

Otherwise, first swap one bit, reverse the back of the substring.

1 /**2 * @param {number[]} nums3 * @return {void} does not return anything, modify Nums in-place instead.4  */5 varNextpermutation =function(nums) {6      for(vari = nums.length-1; I > 0 && nums[i] <= nums[i-1]; i--);7     if(i = = 0){8Reverse (0, Nums.length-1);9         return;Ten     } One      for(varj = i + 1; J < Nums.length && Nums[i-1] < nums[j]; J + +); ASwap (i-1, j-1); -Reverse (i, nums.length-1); -     return;  the      -     functionreverse (start, end) { -          while(Start <end) { - swap (start, end); +start++; -end--; +         } A     } at     functionSwap (I, j) { -         varTMP =Nums[i]; -Nums[i] =Nums[j]; -NUMS[J] =tmp; -     } -};

[Leetcode] [JavaScript] Next permutation