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题目:
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.
Note:
For example, given candidate set 2,3,6,7
and target 7
,
A solution set is:
[7]
[2, 2, 3]
思路:递归,不断地尝试余下的值。
package sum; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class CombinationSum { public List<List<Integer>> combinationSum(int[] candidates, int target) { Arrays.sort(candidates); int len = removeDuplicates(candidates); List<List<Integer>> res = new ArrayList<List<Integer>>(); List<Integer> record = new ArrayList<Integer>(); combine(candidates, 0, len, res, record, target); return res; } private void combine(int[] nums, int start, int end, List<List<Integer>> res, List<Integer> record, int target) { for (int i = start; i < end; ++i) { List<Integer> newRecord = new ArrayList<Integer>(record); newRecord.add(nums[i]); int sum = sum(newRecord); int rem = target - sum; if (rem == 0) { res.add(newRecord); } else if (rem > 0 && rem >= nums[i]) { combine(nums, i, end, res, newRecord, target); } else if (rem < 0) { break; } } } private int sum(List<Integer> record) { int sum = 0; for (int i : record) sum += i; return sum; } private int removeDuplicates(int[] nums) { int n = nums.length; int j = 0; for (int i = 1; i < n; ++i) { if (nums[i] != nums[i - 1]) { nums[++j] = nums[i]; } } return j + 1; } public static void main(String[] args) { // TODO Auto-generated method stub int[] nums = { 1, 2, 2, 3, 6, 7 }; CombinationSum c = new CombinationSum(); List<List<Integer>> res = c.combinationSum(nums, 7); for (List<Integer> l : res) { for (int i : l) System.out.print(i + "\t"); System.out.println(); } } }
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原文地址:http://www.cnblogs.com/shuaiwhu/p/5075503.html