利用回溯法解排列组合问题

回溯法简述

回溯法是一种优选的搜索法，又称试探法。按选优条件向前搜索，已达到目标。但当搜索到某一步时，发现原选择并不优或者达不到目标，就退一步重新选择。这种走不通就退回再走的技术称为回溯法。详见之前的文章 回溯法总结。

排列组合例题

以下三道为leetcode上有关排列组合的问题，分别是46、47和48题。

第一题：46. Permutations

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

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        backtrack(list, new ArrayList<>(), nums);
        return list;
    }
    private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums){
        if(tempList.size() == nums.length){
            list.add(new ArrayList<>(tempList));
        } else{
            for(int i = 0; i < nums.length; i++){
                if(tempList.contains(nums[i])) continue; // element already exists, skip
                tempList.add(nums[i]);
                backtrack(list, tempList, nums);
                tempList.remove(tempList.size() - 1);
            }
        }
    }
}

第二题：47. Permutations II

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

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        Set<List<Integer>> list = new HashSet<>();
        boolean[] used = new boolean[nums.length];
        backtrack(list, new ArrayList<>(), nums, used);
        return new ArrayList<>(list);
    }

    private void backtrack(Set<List<Integer>> list, List<Integer> tempList, int[] nums, boolean[] used) {
        if (tempList.size() == nums.length) {
            list.add(new ArrayList<>(tempList));
        } else {
            for (int i = 0; i < nums.length; i++) {
                if (used[i]) continue;
                used[i] = true;
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, used);
                used[i] = false;
                tempList.remove(tempList.size() - 1);
            }
        }
    }
}

第三题：78. Subsets

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, 0);
        return list;
    }

    private void backtrack(List<List<Integer>> list , List<Integer> tempList, int [] nums, int start){
        list.add(new ArrayList<>(tempList));
        for(int i = start; i < nums.length; i++){
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, i + 1);
            tempList.remove(tempList.size() - 1);
        }
    }
}