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  • LeetCode

    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:

    • All numbers (including target) will be positive integers.
    • Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1a2 ≤ … ≤ ak).
    • 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]

    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 + "	");
                System.out.println();
            }
        }
    
    }
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  • 原文地址:https://www.cnblogs.com/null00/p/5075503.html
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