Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example,
If n = 4 and k = 2, a solution is:
[ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]
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Analysis:
We need consider the boundary and the return condition very carefully!
Iterative Solution:
public class Solution { public List<List<Integer>> combine(int n, int k) { List<List<Integer>> resList = new LinkedList<List<Integer>>(); if (k==0) return resList; LinkedList<Integer> numList = new LinkedList<Integer>(); numList.addLast(0); while (!numList.isEmpty()){ // increase the value of the last number int newLastNum = numList.peekLast()+1; if (newLastNum >= n-(k-numList.size())+1){ // backtrack to the previous level. numList.pollLast(); } else { numList.pollLast(); numList.addLast(newLastNum); // Check if we have k numbers, then output current combination. if (numList.size()==k){ resList.add(new LinkedList<Integer>(numList)); } else { // push a new number into the list. // The initial value is @newLastNum, as we always increase first at the begnning of every loop. numList.addLast(newLastNum); } } } return resList; } }
Solution:
1 public class Solution { 2 public List<List<Integer>> combine(int n, int k) { 3 List<List<Integer>> resSet = new ArrayList<List<Integer>>(); 4 if (n==0 || k==0) return resSet; 5 List<Integer> curList = new ArrayList<Integer>(); 6 combineRecur(resSet,curList,1,k,n,k); 7 return resSet; 8 } 9 10 public void combineRecur(List<List<Integer>> resSet, List<Integer> curList, int curIndex, int numLeft, int n, int k){ 11 if (numLeft==1){ 12 for (int i=curIndex;i<=n;i++){ 13 curList.add(i); 14 List<Integer> temp = new ArrayList<Integer>(); 15 temp.addAll(curList); 16 resSet.add(temp); 17 curList.remove(curList.size()-1); 18 } 19 return; 20 } 21 22 23 for (int i=curIndex;i<=n-numLeft+1;i++){ 24 curList.add(i); 25 combineRecur(resSet,curList,i+1,numLeft-1,n,k); 26 curList.remove(curList.size()-1); 27 } 28 } 29 30 }