LeetCode – Refresh – Combination Sum II-优快云博客

本文介绍了一种解决组合求和问题的方法，特别关注如何通过排序和跳过重复数字来避免结果重复。使用C++实现了一个递归函数getComb，该函数通过检查每个元素是否与前一个元素相同来跳过重复项，并返回所有可能的组合。

The different between I and II is that:

1. For II, each number only can be chosen ONCE.

2. The a number, in the array, encountered more than twice will cause result duplication. So either sort the array and skip the number if it same as the previous, or use find function check whether the result exists in the result vector or not.

For here, sort the input is better, since the output need to be in ascending order.

 1 class Solution {
 2 public:
 3     void getComb(vector<vector<int> > &result, const vector<int> &num, vector<int> current, int sum, int target, int start) {
 4         if (sum > target) {
 5             return;
 6         }
 7         if (sum == target) {
 8             result.push_back(current);
 9             return;
10         }
11         for (int i = start; i < num.size(); i++) {
12             if (i > start && num[i] == num[i-1]) continue;
13             current.push_back(num[i]);
14             getComb(result, num, current, sum + num[i], target, i+1);
15             current.pop_back();
16         }
17     }
18     vector<vector<int> > combinationSum2(vector<int> &num, int target) {
19         vector<vector<int> > result;
20         sort(num.begin(), num.end());
21         getComb(result, num, vector<int> (), 0, target, 0);
22         return result;
23     }
24 };