LeetCode:Combination Sum I II
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]
简单的回溯法(递归实现).
比如对于数组3,2,6,7,target = 7,对数组排序得到[2,3,6,7]
1、第1个数字选取2, 那么接下来就是解决从数组[2,3,6,7]选择数字且target = 7-2 = 5
2、第2个数字选择2,那么接下来就是解决从数组[2,3,6,7]选择数字且target = 5-2 = 3
3、第3个数字选择2,那么接下来就是解决从数组[2,3,6,7]选择数字且target = 3-2 = 1
4、此时target = 1小于数组中的所有数字,失败,回溯,重新选择第3个数字
5、第3个数字选择3,那么接下来就是解决从数组[2,3,6,7]选择数字且target = 3-3 = 0
6、target = 0,找到了一组解,继续回溯寻找其他解
需要注意的是:如果数组中包含重复元素,我们要忽略(因为每个数字可以选择多次,如果不忽略的话,就会产生重复的结果)。貌似oj的测试集数组中都不包含重复的数字
- class Solution {
- private:
- vector<vector<int> > res;
- public:
- vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
- sort(candidates.begin(), candidates.end());//为了输出结果递增,因此先对数组排序
- vector<int> tmpres;
- helper(candidates, 0, target, tmpres);
- return res;
- }
- //从数组candidates[index,...]寻找和为target的组合
- void helper(vector<int> &candidates, const int index, const int target, vector<int>&tmpres)
- {
- if(target == 0)
- {
- res.push_back(tmpres);
- return;
- }
- for(int i = index; i < candidates.size() && target >= candidates[i]; i++)
- if(i == 0 || candidates[i] != candidates[i-1])//由于每个数可以选取多次,因此数组中重复的数就不用考虑
- {
- tmpres.push_back(candidates[i]);
- helper(candidates, i, target - candidates[i], tmpres);
- tmpres.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]
和上一题差不多,只是每个元素只能选一次。
由于有重复元素的存在,比如数组为[1(1),1(2),2,3],target = 6. 可能出现重复结果1(1),2,3 和 1(2),2,3 本文地址
我们可以如下处理:如果数组中当前的数字出现重复,在前面重复了k次,且临时结果数组中也包含了k个当前数字,那么当前的数字可以选择;否则就不选择当前数字
- class Solution {
- private:
- vector<vector<int> >res;
- public:
- vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
- sort(candidates.begin(), candidates.end());
- vector<int> tmpres;
- helper(candidates, 0, target, tmpres, 0);
- return res;
- }
- //从数组candidates[index,...]寻找和为target的组合,times为前一个数字candidates[index-1]重复出现的次数
- void helper(vector<int> &candidates, const int index, const int target, vector<int>&tmpres, int times)
- {
- if(target == 0)
- {
- res.push_back(tmpres);
- return;
- }
- for(int i = index; i < candidates.size() && target >= candidates[i]; i++)
- {
- if(i > 0 && candidates[i] == candidates[i-1])times++;
- else times = 1;
- if(times == 1 || (tmpres.size() >= times-1 && tmpres[tmpres.size()-times+1] == candidates[i]))
- {
- tmpres.push_back(candidates[i]);
- helper(candidates, i+1, target - candidates[i], tmpres, times);
- tmpres.pop_back();
- }
- }
- }
- };
还有一种方法是,在每个子问题的数组中,重复的数字都不选择,这种更简洁
- class Solution {
- private:
- vector<vector<int> >res;
- public:
- vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
- sort(candidates.begin(), candidates.end());
- vector<int> tmpres;
- helper(candidates, 0, target, tmpres);
- return res;
- }
- //从数组candidates[index,...]寻找和为target的组合
- void helper(vector<int> &candidates, const int index, const int target, vector<int>&tmpres)
- {
- if(target == 0)
- {
- res.push_back(tmpres);
- return;
- }
- for(int i = index; i < candidates.size() && target >= candidates[i]; i++)
- {
- if(i > index && candidates[i] == candidates[i-1])continue;//当前子问题中,重复数字都不选择
- tmpres.push_back(candidates[i]);
- helper(candidates, i+1, target - candidates[i], tmpres);
- tmpres.pop_back();
- }
- }
- };
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