topic:
Given a sequence without repeated numbers, 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]
]
Ideas:
Backtracking algorithm
Code:
/**
* @param {number[]} nums
* @return {number[][]}
*/
var permute = function(nums) {
let res = [], track = [];
backtrack(res, nums, track);
return res;
};
var backtrack = function(res, nums, track) {
// 触发结束条件
if (track.length == nums.length) {
return res.push([...track]);;
}
for (let i = 0; i < nums.length; i++) {
// 排除不合法的选择
if (track.includes(nums[i])) continue;
// 做选择
track.push(nums[i]);
// 进入下一层决策树
backtrack(res, nums, track);
// 取消选择
track.pop();
}
}