标签：blog io ar for div sp log on c
这两道题的基本思路和combination那一题是一致的，也是分治的方法。
其中combination Sum复杂一点，因为每个数可能用多次。仔细分析下，本质上也是一样的。原来是每个数仅两种可能。现在每个数有k +１中可能，k = target / i。
所以就是把简单的if else 分支变成for循环：
vector<vector<int> > combHelper(vector<int>::iterator first,
vector<int>::iterator last, int target){
vector<vector<int>> result;
if (target <= 0 || last == first) return result;
if (*(last - 1) == target){
vector<int> vi;
vi.push_back(target);
result.push_back(vi);
auto r2 = combHelper(first, last - 1, target);
for (auto it = r2.begin(); it != r2.end(); it++)
result.push_back(*it);
}
else if (*(last - 1) < target){
auto r1 = combHelper(first, last - 1, target - *(last - 1));
for (auto it = r1.begin(); it != r1.end(); it++){
it->push_back(*(last - 1));
result.push_back(*it);
}
auto r2 = combHelper(first, last - 1, target);
for (auto it = r2.begin(); it != r2.end(); it++)
result.push_back(*it);
}
else {
auto r2 = combHelper(first, last - 1, target);
for (auto it = r2.begin(); it != r2.end(); it++)
result.push_back(*it);
}
return result;
}
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
sort(num.begin(), num.end());
auto result = combHelper(num.begin(), num.end(), target);
if (!result.empty()){
sort(result.begin(), result.end());
auto it = unique(result.begin(), result.end());
result.erase(it, result.end());
}
return result;
}
Combination Sum 和Combination Sum II
标签：blog io ar for div sp log on c
原文地址：http://www.cnblogs.com/hustxujinkang/p/3967332.html
