Combination Sum II

描述

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:

• All numbers (including target) will be positive integers.
• Elements in a combination ($a_1, a_2, ..., a_k$) must be in non-descending order. (ie, $a_1 > a_2 > ... > a_k$).
• The solution set must not contain duplicate combinations.

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]

分析

无

代码

Python

// Combination Sum II
// 时间复杂度O(n!)，空间复杂度O(n)
public class Solution {
    public List<List<Integer>> combinationSum2(int[] nums, int target) {
        Arrays.sort(nums); // 跟第 50 行配合，
        // 确保每个元素最多只用一次
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        dfs(nums, path, result, target, 0);
        return result;
    }
    // 使用nums[start, nums.size())之间的元素，能找到的所有可行解
    private static void dfs(int[] nums, List<Integer> path,
                            List<List<Integer>> result, int gap, int start) {
        if (gap == 0) {  //  找到一个合法解
            result.add(new ArrayList<>(path));
            return;
        }

        int previous = -1;
        for (int i = start; i < nums.length; i++) {
            // 如果上一轮循环已经使用了nums[i]，则本次循环就不能再选nums[i]，
            // 确保nums[i]最多只用一次
            if (previous == nums[i]) continue;

            if (gap < nums[i]) return;  // 剪枝

            previous = nums[i];

            path.add(nums[i]);
            dfs(nums, path, result, gap - nums[i], i + 1);
            path.remove(path.size() - 1);  // 恢复环境
        }
    }
}

Java

// Combination Sum II
// 时间复杂度O(n!)，空间复杂度O(n)
class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &nums, int target) {
        sort(nums.begin(), nums.end()); // 跟第 50 行配合，
                                        // 确保每个元素最多只用一次
        vector<vector<int> > result;
        vector<int> path;
        dfs(nums, path, result, target, 0);
        return result;
    }
private:
    // 使用nums[start, nums.size())之间的元素，能找到的所有可行解
    static void dfs(const vector<int> &nums, vector<int> &path,
            vector<vector<int> > &result, int gap, int start) {
        if (gap == 0) {  //  找到一个合法解
            result.push_back(path);
            return;
        }

        int previous = -1;
        for (size_t i = start; i < nums.size(); i++) {
            // 如果上一轮循环已经使用了nums[i]，则本次循环就不能再选nums[i]，
            // 确保nums[i]最多只用一次
            if (previous == nums[i]) continue;

            if (gap < nums[i]) return;  // 剪枝

            previous = nums[i];

            path.push_back(nums[i]);
            dfs(nums, path, result, gap - nums[i], i + 1);
            path.pop_back();  // 恢复环境
        }
    }
};

C++

# Combination Sum II
# time complexity O(n!), space complexity O(n)
class Solution:
    def combinationSum2(self, nums: list[int], target: int) -> list[list[int]]:
        nums.sort() # works with line 32,
        # ensures each element is used at most once
        result = []
        path = []
        self.dfs(nums, path, result, target, 0)
        return result

    # Use elements between nums[start, nums.size()) to find all possible solutions
    def dfs(self, nums: list[int], path: list[int], result: list[list[int]], gap: int, start: int) -> None:
        if gap == 0: # found a valid solution
            result.append(path[:])
            return

        previous = -1
        for i in range(start, len(nums)):
            # if nums[i] was used in the previous round, we can't use nums[i] in this round,
            # ensures nums[i] is used at most once
            if previous == nums[i]:
                continue

            if gap < nums[i]:
                return # pruning

            previous = nums[i]

            path.append(nums[i])
            self.dfs(nums, path, result, gap - nums[i], i + 1)
            path.pop() # restore environment

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