Combinations
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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],]
枚举出所有的长度为k的子集,一般都用递归+深度搜索
public class Solution { public List<List<Integer>> combine(int n, int k) { List<List<Integer>> solution = new ArrayList<List<Integer>>(); List<Integer> path = new ArrayList<Integer>(); helper(solution,path,1,n,k); return solution; } public void helper(List<List<Integer>> solu,List<Integer> path,int from,int n, int k){if(path.size() == k){List<Integer> lst = new ArrayList<Integer>(path);solu.add(lst);return;}else {for(int i = from;i <= n;i++){path.add(i);helper(solu,path,i+1,n,k);path.remove(path.size()-1);}}}}但是效率有点低的，只打败了15%，能不能进行优化呢？于是我们增加一些条件
public class Solution { public List<List<Integer>> combine(int n, int k) { List<List<Integer>> solution = new ArrayList<List<Integer>>(); if(k <= n && k >= 1){ List<Integer> path = new ArrayList<Integer>(); helper(solution,path,1,n,k); } return solution; } public void helper(List<List<Integer>> solu,List<Integer> path,int from,int n, int k){if(path.size() == k){List<Integer> lst = new ArrayList<Integer>(path);solu.add(lst);}else {for(int i = from;i <= n;i++){path.add(i);helper(solu,path,i+1,n,k);path.remove(path.size()-1);}}}}
此时打败了76.46%0 0
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