leetcode-39-Combination-Sum

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描述

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.

  • The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,

A solution set is: 

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[
  [7],
  [2, 2, 3]
]

分析

类似于八皇后问题，用的是回溯法。思路是先排序，每次递归把剩下的元素一一加到结果集合中，把剩下的元素放到下一层递归去解决子问题，类似于一个深度优先搜索的过程，这类题目的套路都是一样的。

解决方案1(C++)

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class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> result;
        vector<int> path;
        dfs(candidates, path, result, target, 0);
        return result;
    }
    
private:
    void dfs(vector<int>& candidates, vector<int>& path, vector<vector<int>>& result, int target, int start) {
        if(target == 0) {
            result.push_back(path);
            return;
        }
        
        for(int i = start; i < candidates.size(); i++) {
            if(candidates[i] > target) {
                return;
            }
            path.push_back(candidates[i]);
            dfs(candidates, path, result, target-candidates[i], i);
            path.pop_back();
        }
    }
};

解决方案2(Python)

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class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        candidates.sort()
        result = []
        path = []
        self.dfs(candidates, path, result, target)
        return result
        
    def dfs(self, candidates, path, result, target):
        if target == 0:
            result.append(path[:])
            return
        for item, value in enumerate(candidates):
            if target < value:
                return 
            path.append(value)
            self.dfs(candidates[item:], path, result, target-value)
            path.pop()

解决方案3(Java)

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class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(candidates);
        backtrack(candidates, result, new ArrayList<>(), 0, target);
        return result;
    }
    
    private void backtrack(int[] candidates, List<List<Integer>> result, List<Integer> nowList, int left, int target) {
        if (target == 0) {
            result.add(new ArrayList<>(nowList));
            return;
        }
        for (int i = left; i < candidates.length; i++) {
            if (target < candidates[i]) {
                return;
            }
            nowList.add(candidates[i]);
            backtrack(candidates, result, nowList, i, target-candidates[i]);
            nowList.remove(nowList.size()-1);
        }
    }
}

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