[Leetcode] (python): 040-combination Sum II

Source of the topic

https://leetcode.com/problems/combination-sum-ii/

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 is used once in the combination.

Test instructions Analysis

Input:a list as candidates, a value named target

Output:the list number, sumed to target

Conditions: Find a number in the list, make and for target, note that each number can be taken once

Note:

• All numbers (including target) would be positive integers.
• Elements in a combination (a1, a 2, ..., aK) must is in non-descending order. (ie, a1≤ a2≤ ... ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set and 10,1,2,7,6,1,5 target 8 ,
A Solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

Topic ideas

Similar to the previous question, the list is sorted first, then the poor lift, note that there may be duplication, so you need to go to the weight (plus a judgment statement), but also pay attention to the scope of the parameters passed

AC Code (PYTHON)

1 _author_ = "YE" 2 #-*-Coding:utf-8-*-3 4 classsolution (Object): 5 def find (self,candidates, Target, start, valueList): 6 if target = = 0:7 If ValueList not in solution.ans:8 Solution.ans.append (valueList) 9 length =  Len (candidates) Ten for I in range (start, length): One-candidates[i] > target:12 return13 self.find (candidates, Target-candidates[i], i + 1, valueList + [candidates[i]]) + def combinationSum2 (self, candidates, target): 16 "" "17:type candidates:list[int]18:type target:int19:rtype:list[list[int]]20" "" Candidates.sort () Solutio N.ans = []23 self.find (candidates, target, 0, []) solution.ans25 = [candidates] c14>]27 target = 828 s = solution () print (s.combinationsum2 (candidates,target)) /c10> 

[Leetcode] (python): 040-combination Sum II