46. Permutations

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]]

 

和  https://www.cnblogs.com/MarkLeeBYR/p/16887119.html 数组的子集解法相同

本题和78题不一样，递归传入的是index+1，78题递归传入的是j+1。递归路径：

helper(0) -> j=0,helper(1) -> j=1,helper(2) -> j=2,helper(3) -> j=3
          -> j=2,helper(2) -> j=2,helper(3) -> j=3
          -> j=3
          -> j=1,helper(1) -> j=1,helper(2) -> j=2,helper(3) -> j=3
          -> j=2,helper(2) -> j=2,helper(3) -> j=3
          -> j=3
          -> j=2,helper(1) -> j=1,helper(2) -> j=2,helper(3) -> j=3
          -> j=2,helper(2) -> j=2,helper(3) -> j=3
          -> j=3
          -> j=3

public List<List<Integer>> permute(int[] nums) {
    List<List<Integer>> res = new ArrayList<>();
    helper(res, nums, 0);
    return res;
}

private void helper(List<List<Integer>> res, int[] nums, int index) {
    if (index == nums.length) {
        List<Integer> temp = new ArrayList<>();
        for (int num : nums) {
            temp.add(num);
        }
        res.add(temp);
        return;
    }
    for (int j = index; j < nums.length; j++) {
        swap(nums, j, index);
        helper(res, nums, index + 1);
        swap(nums, j, index);
    }
}

private void swap(int[] nums, int m, int n) {
    int temp = nums[m];
    nums[m] = nums[n];
    nums[n] = temp;
}