[leetcode]39combinationsum回溯法找几个数的和为目标值

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * 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]
 ]
 回溯法,每次递归传入的目标值是当前目标值减去此次添加的值，递归函数每次先检测传入的目标值是否是0
 如果是就添加到list中，由于没有负数存在，直接return就行。如果不是，那就遍历数组，要设定一个index记录当前遍历到
 那个数了，递归中的遍历都是从index开始，这样可以避免重复数组的出现。当现在遍历到的数比当前目标值小就递归，
 每次递归结束都要回溯（list删除上一个数）；如果比目标值大了，由于数组已经排序，所以可以直接break。
 */
public class Q39CombinationSum {
    public static void main(String[] args) {
        int[] nums = new int[]{41,24,30,47,40,27,21,20,46,48,23,44,25,49,35,42,36,28,33,32,29,22,37,34,26,45};
        System.out.println(combinationSum(nums,53));
    }
    public static List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        backTracking(result,candidates,new ArrayList<>(),target,0);
        return result;
    }
    public static void backTracking(List<List<Integer>> result,int[] candidates,
                             List<Integer> cur,int left,int index)
    {
        if(left == 0)
        {
            result.add(new ArrayList<>(cur));
            return;
        }
        for(int i = index;i < candidates.length;i++)
        {
            if (candidates[i] <= left)
            {
                cur.add(candidates[i]);
                backTracking(result,candidates,cur,left-candidates[i],i);
                cur.remove(cur.size()-1);
            }
            else
                break;
        }

    }

}