【LeetCode】39. Combination Sum

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39. 组合总和。主题：数组、回溯。难度：中等。

1.题目描述

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

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

给定一个无重复元素的数组candidates和一个目标数target，找出candidates中所有可以使数字和为target的组合。

candidates中的数字可以无限制重复被选取。

Note:

  • All numbers (including target) will be positive integers.

  • The solution set must not contain duplicate combinations.

说明：

  • 所有数字（包括 target）都是正整数。
  • 解集不能包含重复的组合。

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

2.Solutions

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public static List<List<Integer>> combinationSum(int[] nums, int target) {
    List<List<Integer>> list = new ArrayList<>();
    Arrays.sort(nums);
    backtrack(list, new ArrayList<Integer>(), nums, target, 0);
    return list;
}
//典型的回溯处理方法
public static 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{
        //枚举i所有可能的路径
        for(int i = start; i < nums.length; i++){
            tempList.add(nums[i]);
            //not i + 1 because we can reuse same elements
            backtrack(list, tempList, nums, remain - nums[i], i);
            tempList.remove(tempList.size() - 1);//回溯前的清理工作
        }
    }
}

3.举一反三

This structure might apply to many other backtracking questions, but here I am just going to demonstrate Subsets(子集), Permutations(全排列), Combination Sum(组合总和) and Palindrome Partitioning(分割回文串).

Subsets : https://leetcode.com/problems/subsets/

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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/

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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/

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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/

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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/

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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/

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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/

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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;
}
（完）
谢谢你请我吃糖果！
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