daily leetcode - combination-sum - !
题目地址
https://leetcode.com/problems/combination-sum
题目描述
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]
]
思路
像这种结果要求返回所有符合要求解的题十有八九都是要利用到递归,而且解题的思路都大同小异,相类似的题目有 Path Sum II,Subsets II,Permutations,Permutations II,Combinations 等等,如果仔细研究这些题目发现都是一个套路,都是需要另写一个递归函数,这里我们新加入三个变量,start 记录当前的递归到的下标,out 为一个解,res 保存所有已经得到的解,每次调用新的递归函数时,此时的 target 要减去当前数组的的数,具体看代码如下:
解法一:
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> out;
combinationSumDFS(candidates, target, 0, out, res);
return res;
}
void combinationSumDFS(vector<int>& candidates, int target, int start, vector<int>& out, vector<vector<int>>& res) {
if (target < 0) return;
if (target == 0) {res.push_back(out); return;}
for (int i = start; i < candidates.size(); ++i) {
out.push_back(candidates[i]);
combinationSumDFS(candidates, target - candidates[i], i, out, res);
out.pop_back();
}
}
};
我们也可以不使用额外的函数,就在一个函数中完成递归,还是要先给数组排序,然后遍历,如果当前数字大于 target,说明肯定无法组成 target,由于排过序,之后的也无法组成 target,直接 break 掉。如果当前数字正好等于 target,则当前单个数字就是一个解,组成一个数组然后放到结果 res 中。然后将当前位置之后的数组取出来,调用递归函数,注意此时的 target 要减去当前的数字,然后遍历递归结果返回的二维数组,将当前数字加到每一个数组最前面,然后再将每个数组加入结果 res 即可,参见代码如下:
解法二:
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;
}
};
我们也可以用迭代的解法来做,建立一个三维数组 dp,这里 dp[i] 表示目标数为 i+1 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,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:
解法三:
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];
}
};
思路 2
这道题目是求集合,并不是求极值
,因此动态规划不是特别切合,因此我们需要考虑别的方法。
这种题目其实有一个通用的解法,就是回溯法。
网上也有大神给出了这种回溯法解题的
通用写法,这里的所有的解法使用通用方法解答。
除了这道题目还有很多其他题目可以用这种通用解法,具体的题目见后方相关题目部分。
我们先来看下通用解法的解题思路,我画了一张图:
通用写法的具体代码见下方代码区。
关键点解析
- 回溯法
- backtrack 解题公式
代码
- 语言支持: JavaScript,Python3
/*
* @lc app=leetcode id=39 lang=javascript
*
* [39] Combination Sum
*
* https://leetcode.com/problems/combination-sum/description/
*
* algorithms
* Medium (46.89%)
* Total Accepted: 326.7K
* Total Submissions: 684.2K
* Testcase Example: '[2,3,6,7]\n7'
*
* 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]
* ]
*
*/
function backtrack(list, tempList, nums, remain, start) {
if (remain < 0) return;
else if (remain === 0) return list.push([...tempList]);
for (let i = start; i < nums.length; i++) {
tempList.push(nums[i]);
backtrack(list, tempList, nums, remain - nums[i], i); // 数字可以重复使用, i + 1代表不可以重复利用
tempList.pop();
}
}
/**
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/
var combinationSum = function(candidates, target) {
const list = [];
backtrack(list, [], candidates.sort((a, b) => a - b), target, 0);
return list;
};
Python3 Code:
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
"""
回溯法,层层递减,得到符合条件的路径就加入结果集中,超出则剪枝;
主要是要注意一些细节,避免重复等;
"""
size = len(candidates)
if size <= 0:
return []
# 先排序,便于后面剪枝
candidates.sort()
path = []
res = []
self._find_path(target, path, res, candidates, 0, size)
return res
def _find_path(self, target, path, res, candidates, begin, size):
"""沿着路径往下走"""
if target == 0:
res.append(path.copy())
else:
for i in range(begin, size):
left_num = target - candidates[i]
# 如果剩余值为负数,说明超过了,剪枝
if left_num < 0:
break
# 否则把当前值加入路径
path.append(candidates[i])
# 为避免重复解,我们把比当前值小的参数也从下一次寻找中剔除,
# 因为根据他们得出的解一定在之前就找到过了
self._find_path(left_num, path, res, candidates, i, size)
# 记得把当前值移出路径,才能进入下一个值的路径
path.pop()
本文参考自:
https://github.com/grandyang/leetcode/ &
https://github.com/azl397985856/leetcode