【leetcode】46. Permutations 【回溯问题的通用方法】【java】

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]]

This structure might apply to many other backtracking questions, but here I am just going to demonstrate Subsets, Permutations, and Combination Sum.Subsets : https://leetcode.com/problems/subsets/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);    }}
Subsets II (contains duplicates) : https://leetcode.com/problems/subsets-ii/public List<List<Integer>> subsetsWithDup(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++){        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 : https://leetcode.com/problems/permutations/public List<List<Integer>> permute(int[] nums) {   List<List<Integer>> list = new ArrayList<>();   // Arrays.sort(nums); // not necessary   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);      }   }} 
Permutations II (contains duplicates) : https://leetcode.com/problems/permutations-ii/public List<List<Integer>> permuteUnique(int[] nums) {    List<List<Integer>> list = new ArrayList<>();    Arrays.sort(nums);    backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]);    return list;}private void backtrack(List<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] || 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);        }    }}Combination Sum : https://leetcode.com/problems/combination-sum/public List<List<Integer>> combinationSum(int[] nums, int target) {    List<List<Integer>> list = new ArrayList<>();    Arrays.sort(nums);    backtrack(list, new ArrayList<>(), nums, target, 0);    return list;}private void backtrack(List<List<Integer>> list, List<Integer> 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<List<Integer>> combinationSum2(int[] nums, int target) {    List<List<Integer>> list = new ArrayList<>();    Arrays.sort(nums);    backtrack(list, new ArrayList<>(), nums, target, 0);    return list;    }private void backtrack(List<List<Integer>> list, List<Integer> 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<List<String>> partition(String s) {   List<List<String>> list = new ArrayList<>();   backtrack(list, new ArrayList<>(), s, 0);   return list;}public void backtrack(List<List<String>> list, List<String> 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;}