39. Combination Sum

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

官方解答如下（类似的均可用回溯方法解决）：

A general approach to backtracking questions in Java (Subsets, Permutations, Combination Sum, Palindrome Partitioning)

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> 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);
    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++){
        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> 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);
        }
    }
}

---

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);
    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++){
            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;
}