[LeetCode] Permutations (Find all permutations of a integer array w/ or w/o duplicates)

本文链接：https://blog.youkuaiyun.com/craiglin1992/article/details/44852249

本文详细介绍了两种不同的排列算法实现方案：一种使用递归构建列表，通过维护当前元素和剩余元素进行排列；另一种采用交换函数的方式，逐层递增当前索引进行排列。此外，还讨论了在包含重复元素的情况下如何避免重复排列。

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/**
Solution 1, using recursion. 
Construct a list of numbers by maintaining what you have now and what is left. 
Every level you take one element from the current "rest" array and append it to "cur"
**/
public class Solution {
    public List<List<Integer>> permute(int[] num) {
        ArrayList<Integer> rest = new ArrayList<Integer>();
        for(int integer: num) {
            rest.add(integer);
        }
        ArrayList<Integer> cur = new ArrayList<Integer>();
        ArrayList all = new ArrayList();
        permuteHelper(rest,cur,all);
        return all;
    }
    public void permuteHelper(ArrayList<Integer> rest, List cur, List all) {
        if(rest.size()==0){
            all.add(cur);
            return;
        }
        for(Integer integer : rest) {
            cur.add(new Integer(integer));
            ArrayList<Integer> temp = new ArrayList<Integer>(rest);
            temp.remove(integer);
            permuteHelper(temp, new ArrayList<Integer>(cur), all);
            cur.remove(integer);
        }
    }
}

/**
Solution 2
Using a swap function. every level you swap the current index with one of the element on you right you increment current index by one.
**/
public List<List<Integer>> permute(int[] num) {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    if (num == null || num.length == 0) {
      return ans;
    }
    permute(num, 0, ans);
    return ans;
}

//helper function
private void permute(int[] num, int index, List<List<Integer>> ans) {
    if (index == num.length) {
      List<Integer> one = new ArrayList<Integer>();
      for (int i : num) {
        one.add(i);
      }
      ans.add(one);
    } else {
      for (int i = index; i < num.length; ++i) {
        swap(num, index, i);
        permute(num, index+1, ans);
        swap(num, index, i); //swap it back
      }
    }
}

private void swap(int[] num, int i, int j) {
    int temp = num[i];
    num[i] = num[j];
    num[j] = temp;
}

如果有duplicate的话
Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

public List<List<Integer>> permuteUnique(int[] num) {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    if (num == null || num.length == 0) {
        return ans;
    }
    permuteUnique(num, 0, ans);
    return ans;
}
private void permuteUnique(int[] num, int index, List<List<Integer>> ans) {
    if (index == num.length) {
      List<Integer> one = new ArrayList<Integer>();
      for (int i : num) {
        one.add(i);
      }
      ans.add(one);
    } else {
      for (int i = index; i < num.length; ++i) {
      //if the current number has alreay been permuted 
      //"already seen in this round"
        if (!containsDups(num, index, i)) { 
          swap(num, index, i);
          permuteUnique(num, index+1, ans);
          swap(num, index, i);
        }
      }
    }
}

private boolean containsDups(int[] num, int start, int end) {
    for (int i = start; i < end; ++i) {
      if (num[i] == num[end]) {
        return true;
      }
    }
    return false;
}

private void swap(int[] num, int i, int j) {
    int temp = num[i];
    num[i] = num[j];
    num[j] = temp;
}