力扣算法题—039组合求和

#include "000库函数.h"

//第一感觉使用回溯比较快
//好激动，第一次使用回溯成功 96ms,37.1M


class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>>Res;
        if (candidates.size() == 0)return Res;            
        vector<int>v;//临时存放解            
        Combin(candidates, Res,v, target, 0, 0);        
        return Res;
    }

    void Combin(vector<int>candidates, vector<vector<int>>&Res, vector<int>&v, int target, int start, int sum) {
        if (sum == target) {
            Res.push_back(v);
            return;
        }
        else if (sum > target)
            return;
        for (int i = start; i < candidates.size(); ++i) {
            v.push_back(candidates[i]);
            Combin(candidates, Res, v, target, i, sum + candidates[i]);
            v.pop_back();//将数字退出
        }
    }

};


//博客答案
//解法一
class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        sort(candidates.begin(), candidates.end());
        for (int i = 0; i < candidates.size(); ++i) {
            if (candidates[i] > target) break;
            if (candidates[i] == target) { res.push_back({ candidates[i] }); break; }
            vector<int> vec = vector<int>(candidates.begin() + i, candidates.end());
            vector<vector<int>> tmp = combinationSum(vec, target - candidates[i]);
            for (auto a : tmp) {
                a.insert(a.begin(), candidates[i]);
                res.push_back(a);
            }
        }
        return res;
    }
};

//解法二
//建立一个三维数组dp，这里dp[i]表示目标数为i的所有解法集合。
//这里的i就从1遍历到target即可，对于每个i，我们都新建一个二维数组cur，
//然后遍历candidates数组，如果遍历到的数字大于i，说明当前及之后的数字都无法组成i，
//直接break掉。否则如果相等，那么把当前数字自己组成一个数组，并且加到cur中。
//否则就遍历dp[i - candidates[j] - 1] 中的所有数组，如果当前数字大于数组的首元素，
//则跳过，因为我们的结果要求是要有序的。否则就将当前数字加入数组的开头，
//并且将数组放入cur之中即可，参见代码如下：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<vector<int>>> dp;
        sort(candidates.begin(), candidates.end());
        for (int i = 1; i <= target; ++i) {
            vector<vector<int>> cur;
            for (int j = 0; j < candidates.size(); ++j) {
                if (candidates[j] > i) break;
                if (candidates[j] == i) { cur.push_back({ candidates[j] }); break; }
                for (auto a : dp[i - candidates[j] - 1]) {
                    if (candidates[j] > a[0]) continue;
                    a.insert(a.begin(), candidates[j]);
                    cur.push_back(a);
                }
            }
            dp.push_back(cur);
        }
        return dp[target - 1];
    }
};

void T039() {
    vector<int> v;    
    vector<vector<int>>Res;
    Solution s;
    v = { 2, 3, 6, 7 };
    Res = s.combinationSum(v, 7);
    for (auto &a : Res) {
        for (auto b : a)
            cout << b << "  ";
        cout << endl;
    }
    cout << endl<<"*********"<< endl;
    v = { 2, 3,5 };
    Res = s.combinationSum(v, 8);
    for (auto &a : Res) {
        for (auto b : a)
            cout << b << "  ";
        cout << endl;
    }
    cout << endl << "*********" << endl;



}