1 #include "000库函数.h"
2
3 //第一感觉使用回溯比较快
4 //好激动，第一次使用回溯成功 96ms,37.1M
5
6
7 class Solution {
8 public:
9 vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
10 vector<vector<int>>Res;
11 if (candidates.size() == 0)return Res;
12 vector<int>v;//临时存放解
13 Combin(candidates, Res,v, target, 0, 0);
14 return Res;
15 }
16
17 void Combin(vector<int>candidates, vector<vector<int>>&Res, vector<int>&v, int target, int start, int sum) {
18 if (sum == target) {
19 Res.push_back(v);
20 return;
21 }
22 else if (sum > target)
23 return;
24 for (int i = start; i < candidates.size(); ++i) {
25 v.push_back(candidates[i]);
26 Combin(candidates, Res, v, target, i, sum + candidates[i]);
27 v.pop_back();//将数字退出
28 }
29 }
30
31 };
32
33
34 //博客答案
35 //解法一
36 class Solution {
37 public:
38 vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
39 vector<vector<int>> res;
40 sort(candidates.begin(), candidates.end());
41 for (int i = 0; i < candidates.size(); ++i) {
42 if (candidates[i] > target) break;
43 if (candidates[i] == target) { res.push_back({ candidates[i] }); break; }
44 vector<int> vec = vector<int>(candidates.begin() + i, candidates.end());
45 vector<vector<int>> tmp = combinationSum(vec, target - candidates[i]);
46 for (auto a : tmp) {
47 a.insert(a.begin(), candidates[i]);
48 res.push_back(a);
49 }
50 }
51 return res;
52 }
53 };
54
55 //解法二
56 //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。
57 //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，
58 //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，
59 //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。
60 //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，
61 //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，
62 //并且将数组放入cur之中即可，参见代码如下：
63
64 class Solution {
65 public:
66 vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
67 vector<vector<vector<int>>> dp;
68 sort(candidates.begin(), candidates.end());
69 for (int i = 1; i <= target; ++i) {
70 vector<vector<int>> cur;
71 for (int j = 0; j < candidates.size(); ++j) {
72 if (candidates[j] > i) break;
73 if (candidates[j] == i) { cur.push_back({ candidates[j] }); break; }
74 for (auto a : dp[i - candidates[j] - 1]) {
75 if (candidates[j] > a[0]) continue;
76 a.insert(a.begin(), candidates[j]);
77 cur.push_back(a);
78 }
79 }
80 dp.push_back(cur);
81 }
82 return dp[target - 1];
83 }
84 };
85
86 void T039() {
87 vector<int> v;
88 vector<vector<int>>Res;
89 Solution s;
90 v = { 2, 3, 6, 7 };
91 Res = s.combinationSum(v, 7);
92 for (auto &a : Res) {
93 for (auto b : a)
94 cout << b << " ";
95 cout << endl;
96 }
97 cout << endl<<"*********"<< endl;
98 v = { 2, 3,5 };
99 Res = s.combinationSum(v, 8);
100 for (auto &a : Res) {
101 for (auto b : a)
102 cout << b << " ";
103 cout << endl;
104 }
105 cout << endl << "*********" << endl;
106
107
108
109 }
