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Given 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.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
given candidate set 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
分析：递归
1 public class Solution { 2 /** 3 * @param candidates: A list of integers 4 * @param target:An integer 5 * @return: A list of lists of integers 6 */ 7 public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) { 8 ArrayList<Integer> list = new ArrayList<Integer>(); 9 ArrayList<ArrayList<Integer>> listsAll = new ArrayList<ArrayList<Integer>>(); 10 Arrays.sort(candidates); 11 helper(0, 0, candidates, target, list, listsAll); 12 return listsAll; 13 } 14 15 public static void helper(int index, int total, int[] candidates, int target, ArrayList<Integer> list, ArrayList<ArrayList<Integer>> listsAll) { 16 if (index >= candidates.length || total >= target) return; 17 list.add(candidates[index]); 18 total += candidates[index]; 19 if (total == target) { 20 listsAll.add(new ArrayList<Integer>(list)); 21 } 22 helper(index, total, candidates, target, list, listsAll); 23 total = total - list.get(list.size() - 1); 24 list.remove(list.size() - 1); 25 helper(index + 1, total, candidates, target, list, listsAll); 26 } 27 }
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.
Given candidate set [10,1,6,7,2,1,5] and target 8,
A solution set is:
[
[1,7],
[1,2,5],
[2,6],
[1,1,6]
]
1 public class Solution { 2 /** 3 * @param num: Given the candidate numbers 4 * @param target: Given the target number 5 * @return: All the combinations that sum to target 6 */ 7 8 public ArrayList<ArrayList<Integer>> combinationSum2(int[] candidates, int target) { 9 ArrayList<Integer> list = new ArrayList<Integer>(); 10 ArrayList<ArrayList<Integer>> listsAll = new ArrayList<ArrayList<Integer>>(); 11 Arrays.sort(candidates); 12 helper(0, 0, candidates, target, list, listsAll); 13 return listsAll; 14 } 15 16 public static void helper(int index, int total, int[] candidates, int target, ArrayList<Integer> list, ArrayList<ArrayList<Integer>> listsAll) { 17 if (index >= candidates.length || total >= target) return; 18 list.add(candidates[index]); 19 total += candidates[index]; 20 if (total == target) { 21 listsAll.add(new ArrayList<Integer>(list)); 22 } 23 helper(index + 1, total, candidates, target, list, listsAll); 24 total = total - list.get(list.size() - 1); 25 list.remove(list.size() - 1); 26 while (index + 1 < candidates.length && candidates[index] == candidates[index + 1]) { 27 index++; 28 } 29 helper(index + 1, total, candidates, target, list, listsAll); 30 } 31 }
参考请注明出处：cnblogs.com/beiyeqingteng/
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5642221.html
