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。
class Solution {List<List<Integer>> res = new ArrayList<>();List<Integer> temp = new ArrayList<>();public List<List<Integer>> permute(int[] nums) {List<Integer> list = new ArrayList<>();backtrace(nums, list);return res;}// 用list记录添加过的下标。public void backtrace(int[] nums, List<Integer> list){if(list.size() == nums.length){res.add(new ArrayList(temp));return;}for(int i = 0; i < nums.length; i++){int num = nums[i];if(!list.contains(num)) {temp.add(num);list.add(num);} else continue;backtrace(nums, list);temp.remove(temp.size() - 1);list.remove(list.size() - 1);}}}
其实可以直接不记录list,使用nums不重复的特性,在temp中判断是否加入过某个元素:
class Solution {List<List<Integer>> res = new ArrayList<>();List<Integer> temp = new ArrayList<>();public List<List<Integer>> permute(int[] nums) {backtrace(nums);return res;}public void backtrace(int[] nums){if(temp.size() == nums.length){res.add(new ArrayList(temp));return;}for(int i = 0; i < nums.length; i++){// 因为nums不重复!int num = nums[i];if(!temp.contains(num)) {temp.add(num);} else continue;backtrace(nums);temp.remove(temp.size() - 1);}}
}