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:
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]
主要是需要注意如果C中包含了相同的元素的处理方法
如果两个数字相同,前一个数字没有用,那么后面相同的数字也不应该使用,如果前面的数字使用了,可以考虑后面的数字的使用
class Solution { public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { //<DFS算法 vector<vector<int> > result; vector<int> combination; sort(num.begin(), num.end()); result.clear(); internalCombinationSum2(num, 0, 0, target, combination, result); return result; } //<这里有一个问题就是如果C中有重复的数据,如果前一个数据能够放入,后面相同的就可以放入 //<如果前面的数据不能放入,那么后面也就不用放入,注意是按照升序排列的 void internalCombinationSum2(vector<int> &num, int start, int sum, int target, vector<int> &combination, vector<vector<int> > &result) { int size = num.size(); if (sum == target) { result.push_back(combination); return; } else if ((start >= size) || (sum > target)) { return; } for (int i = start; i < size; ) { combination.push_back(num[i]); internalCombinationSum2(num, i + 1, sum + num[i], target, combination, result); combination.pop_back(); i++; while (i < size) { if (num[i] == num[i-1]) { ++i; } else { break; } } } } };
LeetCode—*Combination Sum II(DFS算法C数组中有重复值)
原文:http://blog.csdn.net/xietingcandice/article/details/44592821