Permutations Leetcode

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

 

 

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其实这道题和前面的一样。。。只需要判断存在就跳过就可以了。。。。。。。。。。。。。。。然而。。。。。都是泪。。。。。。

public class Solution {
    List<List<Integer>> result;
    List<Integer> current;
    public List<List<Integer>> permute(int[] nums) {
        result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        current = new ArrayList<>();
        helper(nums, 0);
        return result;
    }
    public void helper(int[] nums, int control) {
        if (control >= nums.length) {
            result.add(new ArrayList<>(current));
        }
        for (int i = 0; i < nums.length; i++) {
            if (current.contains(nums[i])) {
                continue;
            }
            current.add(nums[i]);
            helper(nums, control + 1);
            current.remove(current.size() - 1);
        }
    }
}

这道题还有非递归的方法，挺巧妙的。而且比递归的方法要快。

The idea is that first choose one element, and insert the second element to different places.

[1] -> [1, 2], [2, 1] -> [3, 1, 2], [1, 3, 2], [1, 2, 3], [3, 2, 1], [2, 3, 1], [2, 1, 3]

需要注意的是，每次get的时候其实不能get第k个因为remove长度已经变了，只要get第0个就可以了。

public class Solution {
    List<List<Integer>> result;
    List<Integer> current;
    public List<List<Integer>> permute(int[] nums) {
        result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        current = new ArrayList<>();
        current.add(nums[0]);
        result.add(current);
        for (int i = 1; i < nums.length; i++) {
            int bigSize = result.size();
            for(int k = 0; k < bigSize; k++) {
                List<Integer> l = result.get(0);
                int size = l.size();
                for (int j = 0; j <= size; j++) {
                    List<Integer> tmp = new ArrayList<>(l);
                    tmp.add(j, nums[i]);
                    result.add(tmp);
                }
                result.remove(l);
            }
        }
        return result;
    }
}

也可以用新建list来代替用size控制，快了1ms

public class Solution {
    List<List<Integer>> result;
    List<Integer> current;
    public List<List<Integer>> permute(int[] nums) {
        result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        current = new ArrayList<>();
        current.add(nums[0]);
        result.add(current);
        for (int i = 1; i < nums.length; i++) {
            List<List<Integer>> newResult = new ArrayList<>();
            for(int k = 0; k < result.size(); k++) {
                List<Integer> l = result.get(k);
                int size = l.size();
                for (int j = 0; j <= size; j++) {
                    List<Integer> tmp = new ArrayList<>(l);
                    tmp.add(j, nums[i]);
                    newResult.add(tmp);
                }
            }
            result = newResult;
        }
        return result;
    }
}