77. Combinations
by Botao Xiao
Question
Given two integers n and k, return all possible combinations of k numbers out of 1 … n.
Example
Input: n = 4, k = 2
Output:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
Thinking:
- Method1:回溯
class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> result = new ArrayList<>();
if(n < k) return result;
int[] arr = new int[n];
for(int i = 1; i <= n; i++)
arr[i - 1] = i;
backtrace(k, 0, arr, new ArrayList<Integer>(), result);
return result;
}
private static void backtrace(int k, int start, int[] arr, List<Integer> list, List<List<Integer>> result){
if(k == 0) result.add(new ArrayList<Integer>(list));
else{
for(int i = start; i <= arr.length - k; i++){
list.add(arr[i]);
backtrace(k - 1, i + 1, arr, list, result);
list.remove(list.size() - 1);
}
}
}
}
二刷
还是通过回溯方法,一开始使用size方法来判断,但是发现其实增加一个参数count进行递归可以大幅加快速度。
class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> result = new LinkedList<>();
if(k == 0 || n < k) return result;
backtrace(result, new LinkedList<Integer>(), n, k, 1, 0);
return result;
}
private void backtrace(List<List<Integer>> result, List<Integer> temp, int n, int k, int start, int count){
if(count == k){
result.add(new LinkedList<Integer>(temp));
}else{
for(int i = start; i <= n; i++){
temp.add(i);
backtrace(result, temp, n, k, i + 1, count+1);
temp.remove(count);
}
}
}
}
Third time
- Method 1: Need to use comparison to remove redundant recursions.
class Solution { public List<List<Integer>> combine(int n, int k) { List<List<Integer>> result = new ArrayList<>(); dfs(result, k, n, new ArrayList<Integer>(), 1, 0); return result; } private void dfs(List<List<Integer>> result, int k, int n, List<Integer> temp, int index, int count){ if(count == k){ result.add(new ArrayList<Integer>(temp)); }else if(count < k && count + (n - index + 1) >= k){ for(int i = index; i <= n; i++){ temp.add(i); dfs(result, k, n, temp, i + 1, count + 1); temp.remove(count); } } } }
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