[LintCode] 135 Combination Sum 解题报告

Description
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.

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


Example
Given candidate set [2,3,6,7] and target 7, a solution set is:

[7]
[2, 2, 3]


思路
要求所有的解，我们用DFS。
先排个序，然后开始找。
由于一个数可以出现多次，所以往下一层搜的时候，startIndex还是设置在i。
但是注意还是需要判断，如果有两个x，第一个x如果没有被选，我们不能选第二个x。



Code

public class Solution {
    /**
     * @param candidates: A list of integers
     * @param target:An integer
     * @return: A list of lists of integers
     */
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        // write your code here
        
        List<List<Integer>> res = new ArrayList<>();
        if (candidates == null || candidates.length == 0) {
            return res;
        }
        
        Arrays.sort(candidates);
        List<Integer> list = new ArrayList<>();
        helper(candidates, res, list, target, 0);
        return res;
    }
    
    public void helper(int[] candidates, 
                       List<List<Integer>> res, 
                       List<Integer> list, 
                       int target,
                       int startIndex) {
        
        if (target == 0) {
            res.add(new ArrayList<Integer>(list));
            return;
        }
        
        for (int i = startIndex; i < candidates.length; i++) {
            if (i != startIndex && candidates[i] == candidates[i - 1]) {
                continue;
            }
            if (target < candidates[i]) {
                break;
            }
            list.add(candidates[i]);
            helper(candidates, res, list, target - candidates[i], i);
            list.remove(list.size() - 1);
        }
    }
}