Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates =[10,1,2,7,6,1,5], target =8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]
这题和之前的LeetCode Combination Sum很类似，但是要求同一个数字不能重复选，不过candidates里面可能会有重复的数字。所以策略就是在一个for循环里面判断数字是否重复，注意不能在递归里面判断所选数字是否重复，因为递归的时候是可以选相同数字的，比如样例的[1,1,6]。同时递归的时候传入当前下标的下一个下标。
完整代码如下：
class Solution {
public:
void work(vector<vector<int> >& ans, vector<int>& candidates, int idx, vector<int>& sol, int target)
{
if (target == 0) {
ans.push_back(sol);
}
else {
for (int i = idx; i < candidates.size(); i++) {
if (candidates[i] > target)
break;
if (i != idx && candidates[i] == candidates[i – 1])
continue; // i != idx
sol.push_back(candidates[i]);
work(ans, candidates, i + 1, sol, target – candidates[i]); // i + 1
sol.pop_back();
}
}
}
vector<vector<int> > combinationSum2(vector<int>& candidates, int target)
{
vector<vector<int> > ans;
vector<int> sol;
sort(candidates.begin(), candidates.end());
work(ans, candidates, 0, sol, target);
return ans;
}
};本代码提交AC，用时10MS。
二刷。相比于LeetCode Combination Sum，这一题确实需要对输入数组排序，与上述解法类似，代码如下：
class Solution {
public:
void dfs(vector<int>& candidates, int target, vector<vector<int>>& ans, vector<int>& cur, int idx) {
if (target == 0) {
ans.push_back(cur);
return;
}
if (target < 0)return;
int i = idx;
while (i < candidates.size()) {
if (i > idx&&candidates[i] == candidates[i - 1]) {
++i;
continue;
}
target -= candidates[i];
cur.push_back(candidates[i]);
dfs(candidates, target, ans, cur, i + 1);
target += candidates[i];
cur.pop_back();
++i;
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> ans;
vector<int> cur;
sort(candidates.begin(), candidates.end());
dfs(candidates, target, ans, cur, 0);
return ans;
}
};本代码提交AC，用时8MS。