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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], ]
Solution:
It is like permutation(#46 and #47). Thus we still use recurrence to search for combinations.
given a list called member that contains i numbers in current combination:
if i == k, push member into our final result
else
for each num[j] in member, i <= j < n;
1. push num[j] into member.
2. recursively find combination of (num[j+1…..n], k-(i+1) )
For example, given [1,2,3,4] and k = 2,
we start from 1, member is []
push 1 into member. Now member is [1], and we just collect 1 number. We need to collect 1 more. Go to next recurrence.
Our current position is 0, we should loop from position 1 to n-1:
1. position 1 which is 2. put 2 into member and member becomes [1,2]. Now we have 2 numbers. put member into final result.
2. position 2 which is 3. put 3 into member and member becomes [1,3]. Now we have 2 numbers, put member into final result
3. position 3 is 4. put 4 into member and member becomes [1,4]. Put member into final result.
recurrence starting from 1 ends
we start from 2, member is []
push 2 into member. Now member is [2], and we just collect 1 number. We need to collect 1 more. Go to next recurrence.
Our current position is 1, we should loop from position 2 to n-1:
1. position 2 which is 3. put 3 into member and member becomes [2,3]. Now we have 2 numbers. put member into final result.
2. position 3 which is 4. put 4 into member and member becomes [2,4]. Now we have 2 numbers, put member into final result
recurrence starting from 2 ends
we start from 3, member is []
push 3 into member. Now member [3], and we just collect 1 number. We need to collect 1 more. Go to next recurrence
Our current position is 2, we should loop from position 3 to n-1:
1. position 3 which is 4. put 4 into member and member becomes [3,4]. Now we have 2 numbers. put member into final result.
recurrence starting from 3 ends
we start from 4, member is []
push 4 into member. Now member [4], and we just collect 1 number. We need to collect 1 more. Go to next recurrence
Our current position is 3, we should loop from position 4 to n-1, but 4 > 3. Thus we terminate the recurrence from 4. member is not pushed.
Code:
class Solution {
public:
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> res;
for(int i = 1; i <=n ; i ++){
find_combine(i,n, k-1,vector<int>(), res);
}
return res;
}
void find_combine(int i, int n, int left, vector<int> member, vector<vector<int>> &res){
member.push_back(i);
if(left == 0){
res.push_back(member);
return;
}
else{
i++;
while(i<=n){
find_combine(i, n, left - 1, member, res);
i++;
}
}
}
};
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/188316.html原文链接：https://javaforall.net
