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] ]

 此题的思路可参考「数组的子集」：https://www.cnblogs.com/MarkLeeBYR/p/16887119.html

public List<List<Integer>> combinationSum(int[] nums, int target) {
    List<List<Integer>> list = new ArrayList<>();
    List<Integer> temp = new ArrayList<>();
    Arrays.sort(nums);
    backtrack(list, temp, nums, target, 0);
    return list;
}

private void backtrack(List<List<Integer>> list, List<Integer> temp, int [] nums, int remian, int start) {
    if (remian < 0)
        return;
    else if (remian == 0) {
        list.add(new ArrayList<>(temp));
        return;
    }
    for (int i = start; i < nums.length; i++) {
        temp.add(nums[i]);
        backtrack(list, temp, nums, remian - nums[i], i); //不是i加1，因为可以重复的用元素, 和其他回溯问题的区别：不能传start，必须是i。因为不能走“回头路”，比如2,3,5,target=7，选完3之后，第二轮不会再回过头来选2了
        temp.remove(temp.size() - 1);
    }
}