Combination Sum II(和的组合 II)
【难度:Medium】
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
LeetCode 39题的延伸,在39题的基础上增加了数组有重复数字这一情况,并且不允许出现重复的答案组合。
解题思路
有了39题的经验,此题同样可以沿用39题的解法——回溯来解决,要考虑的是如何避免重复。如果使用39题的代码会发现有两个相同的组合[1,7]和[7,1],所以只要解决了去重的问题即可。
c++代码如下:
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> ans;
vector<int> tmp;
if (candidates.empty())
return ans;
quicksort(candidates,0,candidates.size()-1);
if (target < candidates[0])
return ans;
tranverse(ans,candidates,tmp,target,0);
return ans;
}
void quicksort(vector<int>& n, int low, int high) {
if (low >= high)
return;
int i = low;
int j = high;
int k = n[low];
while (i < j) {
while (i < j && n[j] >= k)
j--;
n[i] = n[j];
while (i < j && n[i] <= k)
i++;
n[j] = n[i];
}
n[i] = k;
quicksort(n,low,i-1);
quicksort(n,i+1,high);
}
void tranverse(vector<vector<int>>& ans, vector<int> n, vector<int>& tmp, int t, int index) {
if (t == 0) {
ans.push_back(tmp);
return;
}
for (int i = index; i < n.size(); i++) {
if (t < n[i])
break;
/*去重的判断在于i == index || n[i] != n[i-1]*/
if ((t == n[i] || t >= 2*n[i]) && (i == index || n[i] != n[i-1])) {
tmp.push_back(n[i]);
tranverse(ans,n,tmp,t-n[i],i+1);
tmp.pop_back();
}
}
return;
}
};