(Leetcode)39&40. Combination Sum--Using Backtracking
2016-07-01 13:46
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Problem
39. Combination SumGiven 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.
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]
]
40. 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 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.
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]
]
Analysis
给定一个数组,从中找出一组数来,使其和等于target。数组无序,但都是正整数。I和II不同的是,I数组里没有重复的数,但一个数可以用多次;II数组里有重复,一个数只能用一次。
I和II都要求返回结果中没有重复的解,且每个解中的数都按非递减排好序。
Solution
39. Combination Sumpublic class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> combination = new ArrayList<List<Integer>>(); List<Integer> singleCombination = new ArrayList<Integer>(); Arrays.sort(candidates); backtrackingCombination(combination, singleCombination, candidates, target, 0); return combination; } void backtrackingCombination(List<List<Integer>> combination, List<Integer> singleCombination, int[] candidates, int target, int start){ if(target>0){ for (int i = start; i <candidates.length && candidates[i]<=target; i++) { singleCombination.add(candidates[i]); backtrackingCombination(combination, singleCombination, candidates, target-candidates[i], i); singleCombination.remove(singleCombination.size()-1); } } else if(target==0){ combination.add(new ArrayList<Integer>(singleCombination)); } } }
40. Combination Sum II
public class Solution { public List<List<Integer>> combinationSum2(int[] candidates, int target) { List<List<Integer>> combination = new ArrayList<List<Integer>>(); List<Integer> singleCombination = new ArrayList<Integer>(); Arrays.sort(candidates); backtrackingCombination(combination, singleCombination, candidates, target, 0); return combination; } void backtrackingCombination(List<List<Integer>> combination, List<Integer> singleCombination, int[] candidates, int target, int start){ if(target>0){ for (int i = start; i <candidates.length && candidates[i]<=target; i++) { if( i>start && candidates[i]==candidates[i-1] ) continue; singleCombination.add(candidates[i]); backtrackingCombination(combination, singleCombination, candidates, target-candidates[i], i+1); singleCombination.remove(singleCombination.size()-1); } } else if(target==0){ combination.add(new ArrayList<Integer>(singleCombination)); } } /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub int[] candidates = {10, 1, 2, 7, 6, 1, 5}; int target = 8; List<List<Integer>> combination = new Solution().combinationSum2(candidates, target); for (int i = 0; i < combination.size(); i++) { for (int j = 0; j < combination.get(i).size(); j++) { System.out.print(combination.get(i).get(j)+" "); } System.out.println(); } } }
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