Lintcode 153. Number Combination II
Title description: Given an array num and an integer target. Find the combination of all the numbers in num as the target.
Lintcode 135. A simple variant of the combination of numbers , but it is a little different when removing duplicates.
class Solution {
public:
/**
* @param num: Given the candidate numbers
* @param target: Given the target number
* @return: All the combinations that sum to target
*/
vector<vector<int>> combinationSum2(vector<int> &num, int target) {
vector<vector<int>> result;
if (0 == num.size()) {
return result;
}
sort(num.begin(), num.end());
vector<int> combinations;
helper(num, target, 0, combinations, result);
return result;
}
//递归函数的定义
/**
*找到所有以combination开头的那些和为target的组合
*并丢到result里，其中剩余的需要加入num里的数的和为remianTarget
* 并且下一个可以加入combination中的数至少从num的startIdx开始（这样可以保证每个数可以取多次)
*/
void helper(vector<int> &num, int remianTarget, int startIdx, vector<int> &combinations, vector<vector<int>> &result) {
//递归的出口
if (0 == remianTarget) {
result.push_back(combinations);
return;
}
//递归的拆解（针对num中每个元素）
for (int i = startIdx; i < num.size(); ++i) {
if (remianTarget < num[i]) {
break;
}
// num = [1,1,1,2] target = 4
// 情况1：^ ^ ^
// 情况2： ^ ^ ^
// 需要去除上面这种组合之间的重复
if (num[i] == num[i - 1] && i != startIdx) {
continue;
}
combinations.push_back(num[i]);
helper(num, remianTarget - num[i], i + 1, combinations, result);
combinations.pop_back();
}
}
};