[LeetCode] 40. 组合总和 II
题目链接 : https://leetcode-cn.com/problems/combination-sum-ii/
题目描述:
给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用一次。
说明：
- 所有数字（包括目标数）都是正整数。
- 解集不能包含重复的组合。
示例:
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
所求解集为:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:
[
[1,2,2],
[5]
]
思路
回溯算法
很标准的模板
关注我的知乎专栏,了解更多解题技巧,大家一起加油!
代码:
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
if not candidates:
return []
candidates.sort()
n = len(candidates)
res = []
def backtrack(i, tmp_sum, tmp_list):
if tmp_sum == target:
res.append(tmp_list)
return
for j in range(i, n):
if tmp_sum + candidates[j] > target : break
if j > i and candidates[j] == candidates[j-1]:continue
backtrack(j + 1, tmp_sum + candidates[j], tmp_list + [candidates[j]])
backtrack(0, 0, [])
return res
java
class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> res = new ArrayList<>();
backtrack(res, candidates, target, 0, 0, new ArrayList<Integer>());
return res;
}
private void backtrack(List<List<Integer>> res, int[] candidates, int target, int i, int tmp_sum, ArrayList<Integer> tmp_list) {
if (tmp_sum == target) {
res.add(new ArrayList<>(tmp_list));
return;
}
for (int start = i; start < candidates.length; start++) {
if (tmp_sum + candidates[start] > target) break;
if (start > i && candidates[start] == candidates[start - 1]) continue;
tmp_list.add(candidates[start]);
backtrack(res, candidates, target, start + 1, tmp_sum + candidates[start], tmp_list);
tmp_list.remove(tmp_list.size() - 1);
}
}
}
类似题目还有:
这类题目都是同一类型的,用回溯算法!
其实回溯算法关键在于:不合适就退回上一步
然后通过约束条件, 减少时间复杂度.
大家可以从下面的解法找出一点感觉!
class Solution:
def subsets(self, nums):
if not nums:
return []
res = []
n = len(nums)
def helper(idx, temp_list):
res.append(temp_list)
for i in range(idx, n):
helper(i + 1, temp_list + [nums[i]])
helper(0, [])
return res
class Solution(object):
def subsetsWithDup(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return []
n = len(nums)
res = []
nums.sort()
# 思路1
def helper1(idx, n, temp_list):
if temp_list not in res:
res.append(temp_list)
for i in range(idx, n):
helper1(i + 1, n, temp_list + [nums[i]])
# 思路2
def helper2(idx, n, temp_list):
res.append(temp_list)
for i in range(idx, n):
if i > idx and nums[i] == nums[i - 1]:
continue
helper2(i + 1, n, temp_list + [nums[i]])
helper2(0, n, [])
return res
class Solution(object):
def permute(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return
res = []
n = len(nums)
visited = [0] * n
def helper1(temp_list,length):
if length == n:
res.append(temp_list)
for i in range(n):
if visited[i] :
continue
visited[i] = 1
helper1(temp_list+[nums[i]],length+1)
visited[i] = 0
def helper2(nums,temp_list,length):
if length == n:
res.append(temp_list)
for i in range(len(nums)):
helper2(nums[:i]+nums[i+1:],temp_list+[nums[i]],length+1)
helper1([],0)
return res
class Solution(object):
def permuteUnique(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return []
nums.sort()
n = len(nums)
visited = [0] * n
res = []
def helper1(temp_list, length):
# if length == n and temp_list not in res:
# res.append(temp_list)
if length == n:
res.append(temp_list)
for i in range(n):
if visited[i] or (i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]):
continue
visited[i] = 1
helper1(temp_list + [nums[i]], length + 1)
visited[i] = 0
def helper2(nums, temp_list, length):
if length == n and temp_list not in res:
res.append(temp_list)
for i in range(len(nums)):
helper2(nums[:i] + nums[i + 1:], temp_list + [nums[i]], length + 1)
helper1([],0)
# helper2(nums, [], 0)
return res
class Solution(object):
def combinationSum(self, candidates, target):
"""
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
"""
if not candidates:
return []
if min(candidates) > target:
return []
candidates.sort()
res = []
def helper(candidates, target, temp_list):
if target == 0:
res.append(temp_list)
if target < 0:
return
for i in range(len(candidates)):
if candidates[i] > target:
break
helper(candidates[i:], target - candidates[i], temp_list + [candidates[i]])
helper(candidates,target,[])
return res
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
if not candidates:
return []
candidates.sort()
n = len(candidates)
res = []
def backtrack(i, tmp_sum, tmp_list):
if tmp_sum == target:
res.append(tmp_list)
return
for j in range(i, n):
if tmp_sum + candidates[j] > target : break
if j > i and candidates[j] == candidates[j-1]:continue
backtrack(j + 1, tmp_sum + candidates[j], tmp_list + [candidates[j]])
backtrack(0, 0, [])
return res
