标签：from log Fix span concat ack one UNC rom
Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
The idea is using recursive approach, we can define a recursive function
helper(prefix, suffix, result)
Where the prefix is: for example
[1]
suffix:
[2, 3]
So every time, we take one value from suffix, put into prefix.
The stop logic is that when suffix length is zero, then we push the prefix into result.
if (sfx.length === 0) { result.push(pfx); }
/** * @param {number[]} nums * @return {number[][]} */ var permute = function(nums) { function helper (pfx, sfx, result) { if (sfx.length === 0) { result.push(pfx); } else { const len = sfx.length; for (let i = 0; i < len; i++) { helper( pfx.concat(sfx[i]), sfx.slice(0, i).concat(sfx.slice(i+1, sfx[len])), result ); } } return result; } return helper([], nums, []); }; permute([1,2,3]) /* [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] */
[], [1,2,3] [1], [2,3] [1, 2], [3], [] | [1, 3], [2] ---> [1,2,3] | [1,2,3] [2], [1, 3] [2, 1], [3] | [2,3] [1] --> [2,1,3] | [2,3,1] [3], [1,2] [3,1][2] | [3,2], [1] --> [3,1,2] | [3,2,1]
标签：from log Fix span concat ack one UNC rom
原文地址：https://www.cnblogs.com/Answer1215/p/12008553.html
