[Algorithm] 46. Permutations

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);
}

 

复制代码
/**
 * @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]]

*/
复制代码

 

复制代码
[], [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]
复制代码

 

posted @   Zhentiw  阅读(235)  评论(0编辑  收藏  举报
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