Leetcode40 Combination Sum II

题目地址：
https://leetcode.com/problems/combination-sum-ii/
描述：
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.

Note:
    All numbers (including target) will be positive integers.
    The solution set must not contain duplicate combinations.
分析：
简单做法就是用递归，用python实现了，见代码1。需要注意有注释的那行，保障了not contain duplicate combinations.

又考虑了非递归做法，用cpp实现了（好久不用cpp，就当复习了），见代码2。
代码1：

class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        result = []
        one = []

        def recursion(idx, sum):
            for x in range(idx, len(candidates)):
                if x != idx and candidates[x] == candidates[x - 1]:  # 同一个slot，不放入相同的值
                    continue
                one.append(candidates[x])
                sum = _sum = sum + candidates[x]
                if _sum < target:
                    recursion(x + 1, sum)
                elif sum == target:
                    result.append(one[:])
                t = one.pop()
                sum -= t
                if _sum >= target:
                    break

        recursion(0, 0)
        return result

代码2：
 

class Solution {
public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
		vector<vector<int>> result;
		vector<int> idxs;
		int idx = 0;
		int sum = 0;
		sort(candidates.begin(), candidates.end());
		idxs.push_back(idx);
		sum += candidates[idx];
		do {
			bool is_push = true;
			if (sum >= target) { // 1.继续push的话会超过target
				if (sum == target) {
					vector<int> tmp;
					for (auto iter = idxs.begin(); iter != idxs.end(); iter++)
						tmp.push_back(candidates[*iter]);
					result.push_back(tmp);
				}
				is_push = false;
			}
			idx = idxs.back();  // 2. 没有可以push的数了
			if (idx == candidates.size() - 1)
				is_push = false;

			if (is_push) { // 继续push
				idxs.push_back(idx + 1);
				sum += candidates[idx + 1];
			}
			else { // pop队尾,直到队尾可以修改为合理值
				while (idxs.size()) {
					sum -= candidates[idxs.back()];
					idxs.pop_back();
					if (idxs.size()) {
						idx = idxs.back();
						if (candidates[idx] != candidates.back()) {
							do {
								idx++;
							} while (candidates[idx] == candidates[idx - 1]);
							break;
						}
					}
				}
				if (idxs.size()) {
					sum -= candidates[idxs.back()];
					idxs[idxs.size() - 1] = idx;
					sum += candidates[idx];
				}
			}
		} while (idxs.size());
		return result;
	}
};

 

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