Leetcode 组合总和系列详解
39. Combination Sum
Medium
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates =[2,3,6,7],target =7, A solution set is: [ [7], [2,2,3] ]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
这道题目没有重复的数字，每个数组都可以重复选取，每个数字又都可以选与不选，因此这个和子集那个递归结构相同，可以加一步排序，剪枝操作。
因为可以重复选取，递归的时候层数还是i
class Solution {
public:
vector> res;
vector> combinationSum(vector& candidates, int target) {
sort(candidates.begin(),candidates.end());
vector temp;
helper(candidates,target,temp,0);
return res;
}
void helper(vector& candidates, int target, vector temp, int start){
if(target<0){
return;
}
if(target==0){
res.push_back(temp);
}
for(int i=start;itarget) break;
target-=candidates[i];
temp.push_back(candidates[i]);
helper(candidates, target, temp, i);
target+=candidates[i];
temp.pop_back();
}
}
}; 40. Combination Sum II
Medium
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Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates =[10,1,2,7,6,1,5], target =8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]
这道题目一个元素只能选取一次，而且存在重复元素，所以要标准的去除剪枝方案。在加上一个排序剪枝。
class Solution {
public:
vector> res;
vector> combinationSum2(vector& candidates, int target) {
sort(candidates.begin(),candidates.end());
vector temp;
helper(candidates,target,temp,0);
return res;
}
void helper(vector& candidates, int target, vector temp, int start){
if(target<0){
return;
}
if(target==0){
res.push_back(temp);
}
for(int i=start;itarget) break;
if(i!=start && candidates[i]==candidates[i-1]) continue;
target-=candidates[i];
temp.push_back(candidates[i]);
helper(candidates, target, temp, i+1);
target+=candidates[i];
temp.pop_back();
}
}
}; 216. Combination Sum III
Medium
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Note:
- All numbers will be positive integers.
- The solution set must not contain duplicate combinations.
Example 1:
Input: k = 3, n = 7 Output: [[1,2,4]]
Example 2:
Input: k = 3, n = 9 Output: [[1,2,6], [1,3,5], [2,3,4]]
这道题目数字只有1到9，只能使用一次，且指定数目，需要剪枝搜索。
class Solution {
public:
int nums[10] = {1,2,3,4,5,6,7,8,9};
vector> res;
vector> combinationSum3(int k, int n) {
vector temp;
helper(k,n,0,0, temp);
return res;
}
void helper(int k, int n,int start, int sum, vector temp){
if(sum>n || temp.size()>k) return;
if(sum==n && temp.size()==k){
res.push_back(temp);
return;
}
for(int i=start;i<9;i++){
sum+=nums[i];
temp.push_back(nums[i]);
helper(k,n,i+1,sum,temp);
sum-=nums[i];
temp.pop_back();
}
}
};