Leetcode 216. Combination Sum III

public class Solution {    private static int rounds;         public static void backTrack(int k, int n, int start, int[] nums, List<Integer> tmp, List<List<Integer>> res) {        if (k == 0) return;        else {            for (int i=start; i<=nums.length-k; i++) {                tmp.add(nums[i]);                backTrack(k-1, n-nums[i], i+1, nums, tmp, res);                // when target is 0 and tmp.size() equals to k means we have reached the node and the sum equals to target,                // add the solution the result set                if (tmp.size() == rounds && n-nums[i] == 0) res.add(new ArrayList<>(tmp));                // before return to the parent level, remove the node we just added                 tmp.remove(tmp.size()-1);            }        }    }        public List<List<Integer>> combinationSum3(int k, int n) {        rounds = k;        int[] nums = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9};        List<List<Integer>> res = new ArrayList<>();        backTrack(k, n, 0, nums, new ArrayList<>(), res);        return res;    }}

Combination Sum I

/** * The algorithm's flow chart would be, * Step i.   add candi[i] to the tmp list, * Step ii.  take care of the scenario which reslut set may have candi[i] by, *           if target drops to zero (reach to node), add all elements in tmp list to the result set *           then return and remove the last element in tmp list return to the parent.  * Step iii. in the for loop, i++, now consider the scenario which reslut set doesn't have candi[i-1] */ public class Solution {    public static void backTrack(int start, int target, int[] candi, List<Integer> tmp, List<List<Integer>> res) {        if (target < 0) return;        else if (target == 0) {            res.add(new ArrayList<>(tmp));            return;        }        else {            // using a for loop tow control if the result set has candi[i] or not            // every time i increament by 1 which excludes candi[i-1] from the result set            for (int i=start; i<candi.length; i++) {                tmp.add(candi[i]);                backTrack(i, target-candi[i], candi, tmp, res);                tmp.remove(tmp.size()-1);            }        }    }        public List<List<Integer>> combinationSum(int[] candidates, int target) {        List<List<Integer>> res = new ArrayList<>();        backTrack(0, target, candidates, new ArrayList<>(), res);        return res;    }}

Combination Sum II

public class Solution {    public static void backTrack(int start, int target, int[] candi, List<Integer> tmp, List<List<Integer>> res) {        if (target < 0) return;        else if (target == 0) {            res.add(new ArrayList<>(tmp));            return;        }        else {            for (int i=start; i<candi.length; i++) {                // we've sorted the array, now just check if two nearby elements are equal to avoid duplicates                // note that i != start should always be placed in front of the rest statement                // (*)                if (i != start && candi[i] == candi[i-1]) continue;                 tmp.add(candi[i]);                // (*)                // difference here we cannot apply candi[i] multiple times,                 // update start to i+1 after we added candi[i] to the result set                backTrack(i+1, target-candi[i], candi, tmp, res);                tmp.remove(tmp.size()-1);            }        }    }        public List<List<Integer>> combinationSum2(int[] candidates, int target) {        // because the array may contain duplicates, sort the array first        Arrays.sort(candidates);        List<List<Integer>> res = new ArrayList<>();        backTrack(0, target, candidates, new ArrayList<>(), res);        return res;    }}