lintcode-medium-Permutations

最新推荐文章于 2020-06-19 19:18:12 发布

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最新推荐文章于 2020-06-19 19:18:12 发布

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原文链接：http://www.cnblogs.com/goblinengineer/p/5351890.html

本文介绍了一种不使用递归的方法来生成给定列表的所有排列，并提供了相应的 Java 代码实现。通过迭代的方式遍历所有可能的排列组合，确保了算法的效率和简洁性。

Given a list of numbers, return all possible permutations.

Example

For nums = [1,2,3], the permutations are:

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

Challenge

Do it without recursion.

class Solution {
    /**
     * @param nums: A list of integers.
     * @return: A list of permutations.
     */
    public ArrayList<ArrayList<Integer>> permute(ArrayList<Integer> nums) {
        // write your code here
        
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        
        if(nums == null || nums.size() == 0)
            return result;
        
        ArrayList<Integer> line = new ArrayList<Integer>();
        boolean[] visited = new boolean[nums.size()];
        
        helper(result, line, visited, nums);
        
        return result;
    }
    
    public void helper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> line, boolean[] visited, ArrayList<Integer> nums){
        
        if(line.size() == nums.size()){
            result.add(new ArrayList<Integer>(line));
            return;
        }
        
        for(int i = 0; i < nums.size(); i++){
            if(!visited[i]){
                line.add(nums.get(i));
                visited[i] = true;
                helper(result, line, visited, nums);
                line.remove(line.size() - 1);
                visited[i] = false;
            }
            
        }
        
        return;
    }
    
}