39. Combination Sum

标签：oid   div   color   limit   numbers   chosen   for   set   get   

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

class Solution {
public List<List<Integer>> combinationSum(int[] nums, int target) {
    List<List<Integer>> list = new ArrayList<>();
    backtrack(list, new ArrayList<>(), nums, target, 0);
    return list;
}

private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) list.add(new ArrayList<>(tempList));
    else{ 
        for(int i = start; i < nums.length; i++){
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, remain - nums[i], i); 
            //找到了一个解或者 remain < 0 了，将当前数字移除，然后继续尝试
            tempList.remove(tempList.size() - 1);
        }
    }
}

}

大佬给的答案是回溯法，和我想的一样，但是我只想出了个大概，用code写不出我的想法，所以还要多练。

39. Combination Sum

标签：oid   div   color   limit   numbers   chosen   for   set   get   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/10354380.html