排列组合

77. Combinations

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],]

题解：

和N-Queens都是NP问题，利用helper迭代，迭代的stop condition是item.size() == k, 此时把item的copy加到res中。

若item.size()还没有到k, item加i, 然后继续迭代，到了k返回后去掉item最后一个值。

这里的start是递归的关键，每次recursion时, 加完i后，把i+1赋给start继续recursion.
*/
public class Solution {
    public  List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> rst = new ArrayList<List<Integer>>();
        List<Integer> solution = new ArrayList<Integer>();
        helper(rst, solution, n, k, 1);
        return rst;
    }
            
    private void helper(List<List<Integer>> rst, List<Integer> solution,  int n,  int k, int start) {
        if (solution.size() == k){
            rst.add(new ArrayList(solution));
            //return;
        }
        for(int i = start; i <= n; i++){
            solution.add(i);
            // the new start should be after the next number after i
            helper(rst, solution, n, k, i+1); 
            solution.remove(solution.size() - 1);
        }
    }

}

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[  [1,2,3],  [1,3,2],  [2,1,3],  [2,3,1],  [3,1,2],  [3,2,1]]

public class Solution {
    public List<List<Integer>> permute(int[] nums) {
         ArrayList<List<Integer>> rst = new ArrayList<List<Integer>>();
         if (nums == null || nums.length == 0) {
             return rst; 
         }


         ArrayList<Integer> list = new ArrayList<Integer>();
         helper(rst, list, nums);
         return rst;
    }
    
    public void helper(ArrayList<List<Integer>> rst, ArrayList<Integer> list, int[] nums){
        if(list.size() == nums.length) {
            rst.add(new ArrayList<Integer>(list));
            return;
        }
        
        for(int i = 0; i < nums.length; i++){
            if(list.contains(nums[i])){
                continue;
            }
            list.add(nums[i]);
            helper(rst, list, nums);
            list.remove(list.size() - 1);
        }
    }
}

 Permutations II

Add to List

DescriptionSubmissionsSolutions

Really easy Java solution, much easier than the solutions with very high vote

Use an extra boolean array " boolean[] used" to indicate whether the value is added to list.

Sort the array "int[] nums" to make sure we can skip the same value.

when a number has the same value with its previous, we can use this number only if his previous is used

public class Solution {    public List<List<Integer>> permuteUnique(int[] nums) {        List<List<Integer>> res = new ArrayList<List<Integer>>();        if(nums==null || nums.length==0) return res;        boolean[] used = new boolean[nums.length];        List<Integer> list = new ArrayList<Integer>();        Arrays.sort(nums);        dfs(nums, used, list, res);        return res;    }    public void dfs(int[] nums, boolean[] used, List<Integer> list, List<List<Integer>> res){        if(list.size()==nums.length){            res.add(new ArrayList<Integer>(list));            return;        }        for(int i=0;i<nums.length;i++){            if(used[i]) continue;            if(i>0 &&nums[i-1]==nums[i] && !used[i-1]) continue;            used[i]=true;            list.add(nums[i]);            dfs(nums,used,list,res);            used[i]=false;            list.remove(list.size()-1);        }    }}

方法二：

DescriptionSubmissionsSolutions

Share my Java code with detailed explanantion

public class Solution {    public List<List<Integer>> permuteUnique(int[] nums) {        List<List<Integer>> ans = new ArrayList<>();        if (nums==null || nums.length==0) { return ans; }        permute(ans, nums, 0);        return ans;    }        private void permute(List<List<Integer>> ans, int[] nums, int index) {        if (index == nums.length) {             List<Integer> temp = new ArrayList<>();            for (int num: nums) { temp.add(num); }            ans.add(temp);            return;        }        Set<Integer> appeared = new HashSet<>();        for (int i=index; i<nums.length; ++i) {            if (appeared.add(nums[i])) {                swap(nums, index, i);                permute(ans, nums, index+1);                swap(nums, index, i);            }        }    }        private void swap(int[] nums, int i, int j) {        int save = nums[i];        nums[i] = nums[j];        nums[j] = save;    }}

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