Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
Example 1:
Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Example 2:
Input: nums = [0,1] Output: [[0,1],[1,0]]
Example 3:
Input: nums = [1] Output: [[1]]
Constraints:
1 <= nums.length <= 6-10 <= nums[i] <= 10- All the integers of
numsare unique.
给定一个不含重复数字的数组 nums ,返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
示例 1:
输入:nums = [1,2,3] 输出:[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
示例 2:
输入:nums = [0,1] 输出:[[0,1],[1,0]]
示例 3:
输入:nums = [1] 输出:[[1]]
提示:
1 <= nums.length <= 6-10 <= nums[i] <= 10nums中的所有整数 互不相同
| Language | Runtime | Memory | Submission Time |
|---|---|---|---|
| typescript | 76 ms | 47.1 MB | 2022/05/08 23:47 |
function permute(nums: number[]): number[][] {
if (nums.length === 1) {
return [nums];
}
let res: number[][] = [];
function dfs(idx: number): void {
if (idx === nums.length) {
return;
}
if (idx === 1) {
res = [[nums[0], nums[1]], [nums[1], nums[0]]];
dfs(idx+1);
return;
}
res = res.reduce((acc: number[][], item) => {
const newItem: number[][] = [];
for (let i = 0; i <= item.length; i++) {
newItem.push([
...item.slice(0, i),
nums[idx],
...item.slice(i),
])
}
return [...acc, ...newItem];
}, [])
dfs(idx+1);
}
dfs(1);
return res;
};No notes