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46. Permutations 全排列,無重複

Given a collection ofdistinctintegers, 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]
]

if(temp.contains(nums[i])) continue;temp里加過了的話,就不加了。這句話得有

class Solution {
    public List<List<Integer>> permute(int
[] nums) { //cc List<List<Integer>> results = new ArrayList<List<Integer>>(); if (nums == null || nums.length == 0) return results; //排序一下 Arrays.sort(nums); dfs(nums, new ArrayList<Integer>(), 0, results);
return results; } public void dfs(int[] nums, List<Integer> temp, int start, List<List<Integer>> results) { //exit if (temp.size() == nums.length) results.add(new ArrayList<>(temp)); for (int i = 0; i < nums.length; i++) {
// if ((i > start) && (nums[i] == nums[i - 1])) // continue; if(temp.contains(nums[i])) continue; temp.add(nums[i]); dfs(nums, temp, i + 1, results); temp.remove(temp.size() - 1); } } }
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