Combination Sum (lintcode 135)
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Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Notice
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.
基本思想就是将vector数据进行排序,然后去重,最后利用完全背包问题做DP。
#include <iostream>#include <algorithm>#include <vector>#include <cstdio>using namespace std;class Solution {public: /* * @param candidates: A list of integers * @param target: An integer * @return: A list of lists of integers */ static vector<vector<int>> combinationSum(vector<int> &candidates, int target) { // write your code here std::sort(candidates.begin(), candidates.end()); vector<int>::iterator last_it = std::unique(candidates.begin(), candidates.end()); candidates.resize(std::distance(candidates.begin(), last_it)); vector<vector<int>> result_set; vector<int> unique_result; unique_result.reserve(std::distance(candidates.begin(), last_it)); helper(candidates, 0, target, unique_result, result_set); return result_set; } static void helper(vector<int>& arr, int left_num, int target, vector<int>& unique_result, vector<vector<int>>& result_set) { if (left_num >= 0 && left_num < arr.size()) { if (target == 0) { result_set.push_back(unique_result); } else if (target >= arr[left_num]) { unique_result.push_back(arr[left_num]); helper(arr, left_num, target - arr[left_num], unique_result, result_set); unique_result.pop_back(); helper(arr, left_num + 1, target, unique_result, result_set); } } }};int main(int argc, char * * argv, char * * env){ vector<int> candidates = { 2, 3, 6, 7 }; vector<vector<int>> result_set = Solution::combinationSum(candidates, 7); for (auto& itvec:result_set) { for (auto& it:itvec) { cout << it; } cout << std::endl; } char ch; scanf_s("%c", &ch);}
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