LeetCode--------39. Combination Sum(组合和)-优快云博客

本文链接：https://blog.youkuaiyun.com/nameix/article/details/52743223

本文介绍了一种寻找候选数集合中所有可能的组合，使得这些组合的和等于目标数的算法。该算法允许无限次重复使用集合中的元素，并且确保找到的所有组合都是唯一的。文章通过示例展示了如何实现这一算法。

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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.

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]
]

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public class Solution {
    public  List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (candidates == null) {
            return result;
        }

        List<Integer> path = new ArrayList<Integer>();
        Arrays.sort(candidates);
        helper(candidates, target, path, 0, result);

        return result;
    }

     void helper(int[] candidates, int target, List<Integer> path, int index,
        List<List<Integer>> result) {
        if (target == 0) {
            result.add(new ArrayList<Integer>(path));
            return;
        }

        int prev = -1;
        for (int i = index; i < candidates.length; i++) {
            if (candidates[i] > target) {
                break;
            }

            if (prev != -1 && prev == candidates[i]) {
                continue;
            }

            path.add(candidates[i]);
            helper(candidates, target - candidates[i], path, i, result);
            path.remove(path.size() - 1);

            prev = candidates[i];
        }
    }
}

public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    int[] cans = {};
    
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        this.cans = candidates;
        Arrays.sort(cans);
        backTracking(new ArrayList(), 0, target);
        return ans;
    }
    
    public void backTracking(List<Integer> cur, int from, int target) {
        if (target == 0) {
            List<Integer> list = new ArrayList<Integer>(cur);
            ans.add(list);
        } else {
            for (int i = from; i < cans.length && cans[i] <= target; i++) {
                cur.add(cans[i]);
                backTracking(cur, i, target - cans[i]);
                cur.remove(new Integer(cans[i]));
            }
        }
    }
}