题目:
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.
- 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] ]
解读:给予一组不重复的正整数数组C,和一个正整数T。从C数组中找出所有数字组合使得和为T,所用的元素不可以重复。最终结果中的数组不能重复。
思维:采用递归的方法。设一个整数sum存储当前预备结果的数组的元素和,如果sum>target则返回并把预备结果数组最后一个元素弹出;如果sum==target则把预备结果数组放入最终的结果中,返回并弹出最后元素。用low来存储搜索的数组开始的下标。
注意:所用元素不可以重复 控制low值。
最终结果不能重复,先排序,控制解中的最小值开头的不重复。
代码:
class Solution {
public:
void combinesum(int target, int& sum, vector<int>& candidates,vector<int>& solution,
int low, vector<vector<int>>& result){
if(sum > target) return;
if(sum == target) result.push_back(solution);
for(int i = low; i < candidates.size(); i++){
sum += candidates[i];
solution.push_back(candidates[i]);
combinesum(target, sum, candidates, solution, i+1 , result);
sum -= candidates[i];
solution.pop_back();
while(i < candidates.size()-1&& candidates[i] == candidates[i+1]) i++;
}
}
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
int sum = 0;
int low = 0;
vector<vector<int>> result;
vector<int> solution;
sort(candidates.begin(), candidates.end());
combinesum(target,sum, candidates,solution, low, result);
return result;
}
};