回溯法解决排列组合问题

package 算法和数据结构;
/**
* Filename : Backtracking.java
* Author : [email protected]
* Creation time : 上午10:16:04 - 2017年3月13日
* Description :   利用回溯法来解决诸如 子集数量，排列，组合 的问题。
*/
import java.util.*;

public class Backtracking {

    /**
     * Subsets
     */
    //**********************************************************************
    //Subsets : https://leetcode.com/problems/subsets/
    public List> subsets(int[] nums) {
        List> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, 0);
        return list;
    }

    private void backtrack(List> list , List 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);
        }
    }

    //********************************************************************
    //Subsets II (contains duplicates) : https://leetcode.com/problems/subsets-ii/
    public List> subsetsWithDup(int[] nums) {
        List> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack2(list, new ArrayList<>(), nums, 0);
        return list;
    }

    private void backtrack2(List> list, List tempList, int [] nums, int start){
        list.add(new ArrayList<>(tempList));
        for(int i = start; i < nums.length; i++){
            if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, i + 1);
            tempList.remove(tempList.size() - 1);
        }
    } 


/**
 * Permutations排列
 */
    //***************************************************************************
    //Permutations : https://leetcode.com/problems/permutations/
    public List> permute(int[] nums) {
       List> list = new ArrayList<>();
       // Arrays.sort(nums); // not necessary
       backtrack(list, new ArrayList<>(), nums);
       return list;
    }

    private void backtrack(List> list, List 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);
          }
       }
    } 

    //*******************************************************************************************
    //Permutations II (contains duplicates) : https://leetcode.com/problems/permutations-ii/
    public List> permuteUnique(int[] nums) {
        List> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]);
        return list;
    }

    private void backtrack(List> list, List 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] || i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue;
                used[i] = true; 
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, used);
                used[i] = false; 
                tempList.remove(tempList.size() - 1);
            }
        }
    }

    /**
     * 组合
     */

    //****************************************************************************
    //Combination Sum : https://leetcode.com/problems/combination-sum/
    public List> combinationSum(int[] nums, int target) {
        List> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(list, new ArrayList<>(), nums, target, 0);
        return list;
    }

    private void backtrack(List> list, List tempList, int [] nums, int remain, int start){
        if(remain < 0) return;
        else if(remain == 0) list.add(new ArrayList<>(tempList));
        else{ 
            for(int i = start; i < nums.length; i++){
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, remain - nums[i], i); // not i + 1 because we can reuse same elements
                tempList.remove(tempList.size() - 1);
            }
        }
    }

    //Combination Sum II (can't reuse same element) : https://leetcode.com/problems/combination-sum-ii/
    public List> combinationSum2(int[] nums, int target) {
        List> list = new ArrayList<>();
        Arrays.sort(nums);
        backtrack2(list, new ArrayList<>(), nums, target, 0);
        return list;

    }

    private void backtrack2(List> list, List tempList, int [] nums, int remain, int start){
        if(remain < 0) return;
        else if(remain == 0) list.add(new ArrayList<>(tempList));
        else{
            for(int i = start; i < nums.length; i++){
                if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, remain - nums[i], i + 1);
                tempList.remove(tempList.size() - 1); 
            }
        }
    } 

    //Palindrome Partitioning : https://leetcode.com/problems/palindrome-partitioning/
    public List> partition(String s) {
       List> list = new ArrayList<>();
       backtrack(list, new ArrayList<>(), s, 0);
       return list;
    }

    public void backtrack(List> list, List tempList, String s, int start){
       if(start == s.length())
          list.add(new ArrayList<>(tempList));
       else{
          for(int i = start; i < s.length(); i++){
             if(isPalindrome(s, start, i)){
                tempList.add(s.substring(start, i + 1));
                backtrack(list, tempList, s, i + 1);
                tempList.remove(tempList.size() - 1);
             }
          }
       }
    }

    public boolean isPalindrome(String s, int low, int high){
       while(low < high)
          if(s.charAt(low++) != s.charAt(high--)) return false;
       return true;
    } 

}