(Leetcode)39&40. Combination Sum--Using Backtracking

Problem

39. Combination Sum

Given a set of candidate numbers (C) 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]
]

40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

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

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:

[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Analysis

给定一个数组，从中找出一组数来，使其和等于target。数组无序，但都是正整数。

I和II不同的是，I数组里没有重复的数，但一个数可以用多次；II数组里有重复，一个数只能用一次。

I和II都要求返回结果中没有重复的解，且每个解中的数都按非递减排好序。

Solution

39. Combination Sum

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> combination = new ArrayList<List<Integer>>();
        List<Integer> singleCombination = new ArrayList<Integer>();
        Arrays.sort(candidates);
        backtrackingCombination(combination, singleCombination, candidates, target, 0);

        return combination;
    }


    void backtrackingCombination(List<List<Integer>> combination, List<Integer> singleCombination, int[] candidates, int target, int start){
        if(target>0){
            for (int i = start; i <candidates.length && candidates[i]<=target; i++) {
                singleCombination.add(candidates[i]);
                backtrackingCombination(combination, singleCombination, candidates, target-candidates[i], i);
                singleCombination.remove(singleCombination.size()-1);           
            }

        }

        else if(target==0){
             combination.add(new ArrayList<Integer>(singleCombination));
        }
    }
}

40. Combination Sum II

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> combination = new ArrayList<List<Integer>>();
        List<Integer> singleCombination = new ArrayList<Integer>();
        Arrays.sort(candidates);
        backtrackingCombination(combination, singleCombination, candidates, target, 0);

        return combination;
    }


    void backtrackingCombination(List<List<Integer>> combination, List<Integer> singleCombination, int[] candidates, int target, int start){
        if(target>0){
            for (int i = start; i <candidates.length && candidates[i]<=target; i++) {
                if( i>start && candidates[i]==candidates[i-1] )
                    continue;
                singleCombination.add(candidates[i]);
                backtrackingCombination(combination, singleCombination, candidates, target-candidates[i], i+1);
                singleCombination.remove(singleCombination.size()-1);           
            }

        }

        else if(target==0){
             combination.add(new ArrayList<Integer>(singleCombination));
        }
    }
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] candidates = {10, 1, 2, 7, 6, 1, 5};
        int target = 8;
        List<List<Integer>> combination = new Solution().combinationSum2(candidates, target);
        for (int i = 0; i < combination.size(); i++) {
            for (int j = 0; j < combination.get(i).size(); j++) {
                System.out.print(combination.get(i).get(j)+" ");
            }
            System.out.println();
        }
    }

}