大家好,又见面了,我是全栈君,今天给大家准备了Idea注册码。
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.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- 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]
class Solution { private: vector<vector<int> > ivvec; public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<int> ivec; sort(candidates.begin(), candidates.end()); combinationSum(candidates, 0, target, ivec); return ivvec; } void combinationSum(vector<int> &candidates, int beg, int target, vector<int> ivec) { if (0 == target) { ivvec.push_back(ivec); return; } for (int idx = beg; idx < candidates.size(); ++idx) { if (target - candidates[idx] < 0) break; ivec.push_back(candidates[idx]); combinationSum(candidates, idx, target - candidates[idx], ivec); ivec.pop_back(); } } };
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 (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- 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]
与上题差别不大。依然是用DFS,基本的问题在于怎样去重。
能够添加一个剪枝: 当当前元素跟前一个元素是同样的时候。假设前一个元素被取了,那当前元素能够被取,也能够不取,反过来假设前一个元素没有取。那我们这个以及之后的所以同样元素都不能被取。
(採用flag的向量作为标记元素是否被选取)
class Solution {private: vector<vector<int> > ivvec; vector<bool> flag;public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<int> ivec; sort(num.begin(), num.end()); flag.resize(num.size()); for (int i = 0; i < flag.size(); ++i) flag[i] = false; combinationSum2(num, 0, ivec, target, 0); return ivvec; } void combinationSum2(vector<int> &num, int beg, vector<int> ivec, int target, int sum) { if (sum > target) return; if (sum == target) { ivvec.push_back(ivec); return; } for (int idx = beg; idx < num.size(); ++idx) { if (sum + num[idx] > target) continue; if (idx != 0 && num[idx] == num[idx - 1] && flag[idx - 1] == false) continue; flag[idx] = true; ivec.push_back(num[idx]); combinationSum2(num, idx + 1, ivec, target, sum + num[idx]); ivec.pop_back(); flag[idx] = false; } }};
版权声明:本文博主原创文章。博客,未经同意不得转载。
发布者:全栈程序员-用户IM,转载请注明出处:https://javaforall.cn/117022.html原文链接:https://javaforall.cn
【正版授权,激活自己账号】: Jetbrains全家桶Ide使用,1年售后保障,每天仅需1毛
【官方授权 正版激活】: 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...