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.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

public class Solution {
    ArrayList<ArrayList<Integer>> result = null;
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        result = new ArrayList<ArrayList<Integer>>();
        getSolution(new ArrayList<Integer>(), target, candidates);
        return result;
    }
    public void getSolution(ArrayList<Integer> row, int target, int[] candidates){
        int i = 0;
        if(target == 0){
            result.add(row);
        }
        while(i < candidates.length && candidates[i] <= target){
            row.add(candidates[i]);
            target -= candidates[i];
            getSolution(new ArrayList<Integer>(row),target,candidates);
            row.remove(row.size() - 1);
            target += candidates[i];
            i ++;
        }
        if(i == 0){
            return;
        }
    }
}

没有去重。

eg: [1,1,2] 3

去重包括两部分，1.array 里面有重复： 1，1

                     2.顺序， 如 1,2 ; 2, 1

对于1，先将数组去重。

对于2，后面的选择范围只能从当前选择的元素开始，只能选不比它小的元素。

public class Solution {
    ArrayList<ArrayList<Integer>> result = null;
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        Arrays.sort(candidates);
        int start = 0;
        for(int i = 1; i < candidates.length; i ++){
            if(candidates[i] != candidates[start]){
                candidates[++start] = candidates[i];
            }
        }
        result = new ArrayList<ArrayList<Integer>>();
        getSolution(new ArrayList<Integer>(), target, candidates, start, 0);
        return result;
    }
    public void getSolution(ArrayList<Integer> row, int target, int[] candidates, int len, int pos){
        int i = pos;
        if(target == 0){
            result.add(row);
        }
        while(i <= len && candidates[i] <= target){
            row.add(candidates[i]);
            target -= candidates[i];
            getSolution(new ArrayList<Integer>(row), target, candidates, len, i);
            row.remove(row.size() - 1);
            target += candidates[i];
            i ++;
        }
29     }
30 }

 第三遍：

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        int len = 0;
        Arrays.sort(candidates);
        for(int i = 0; i < candidates.length; i ++){
            if(i == 0 || candidates[i] != candidates[i - 1]){
                candidates[len ++] = candidates[i];
            }
        }
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        findElement(candidates, 0, len, result, new ArrayList<Integer>(), target);
        return result;
    }
    
    public void findElement(int[] arr, int start, int end, List<List<Integer>> result, ArrayList<Integer> row, int tar){
        if(tar == 0) result.add(row);
        for(int i = start; i < end; i ++){
            if(arr[i] <= tar){
                ArrayList<Integer> rown = new ArrayList<Integer> (row);
                rown.add(arr[i]);
                findElement(arr, i, end, result, rown, tar - arr[i]);
            }
        }
    }
}