Leetcode 39. Combination Sum

题目

Given a set of candidate numbers (C) (without duplicates) 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:

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

思路

从第一个往前递归计算

测试用例

[] 3[2] 3[1,2,3,4,5] 8[2] 0

代码

package leetcodeArray;import java.util.ArrayList;import java.util.Arrays;import java.util.LinkedList;import java.util.List;public class Leetcode39CombinationSum {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        List<List<Integer>> result = new ArrayList<List<Integer>>();        List<Integer> cur = new LinkedList<Integer>();;        combination(candidates, target, result, cur, 0);        return result;    }    private void combination(int[] candidate, int target, List<List<Integer>> result, List<Integer> cur, int start){        if(target == 0){            result.add(new LinkedList<Integer>(cur));            return;        }        for(int i = start; i < candidate.length && candidate[i] <= target;i++){            cur.add(candidate[i]);            combination(candidate, target - candidate[i], result, cur, i);            cur.remove(cur.size() - 1);        }    }}

结果

他山之玉

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        List<List<Integer>> result = new ArrayList<>();        dfs(result, new LinkedList<Integer>(), candidates, target);        return result;    }    private void dfs(List<List<Integer>> result, LinkedList<Integer> list, int[] arr, int target) {        if (target == 0) {            result.add(new LinkedList<Integer>(list));            return;        }        for (int i = arr.length - 1; i >= 0; i--) {            if (arr[i] <= target) {                list.addFirst(arr[i]);                dfs(result, list, Arrays.copyOfRange(arr, 0, i + 1), target - arr[i]);                list.removeFirst();            }        }    }}

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

def combinationSum(self, candidates, target):    res = []    candidates.sort()    self.dfs(candidates, target, 0, [], res)    return resdef dfs(self, nums, target, index, path, res):    if target < 0:        return  # backtracking    if target == 0:        res.append(path)        return     for i in xrange(index, len(nums)):        self.dfs(nums, target-nums[i], i, path+[nums[i]], res)