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[leetcode]Combination SumII

Combination Sum II

 

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums toT.

Each number in C may only be used once in the combination.

Note:

  • 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.

 

For example, given candidate set 10,1,2,7,6,1,5 and target 8
A solution set is: 
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

算法思路:

与[leetcode]Combination Sum类似,区别是这里的元素要求不能重复,且结果需要去重。

代码如下:

 1 public class Solution { 2      List<List<Integer>> result = new ArrayList<List<Integer>>(); 3     public List<List<Integer>> combinationSum2(int[] num, int target) { 4         if(num == null || num.length == 0) return result; 5         Arrays.sort(num); 6         List<Integer> list = new ArrayList<Integer>(); 7         dfs(num,list,0,target); 8         return result; 9     }10     11     private void dfs(int[] num,List<Integer> list,int k,int target){12         if(target == 0){13             List<Integer> copy = new ArrayList<Integer>(list);14             for(List<Integer> node : result){15                 int count = 0;16                 if(copy.size() == node.size()){17                     for(int i = 0; i < node.size(); i++){18                         if(copy.get(i) == node.get(i)){19                             count++;20                         }else break;21                     }22                     if(count == node.size()) return;23                 }24             }25             result.add(copy);26             return;27         }28         if(k >= num.length || num[k] > target) return;29         for(int i = k; i < num.length; i++){30             list.add(num[i]);31             dfs(num,list,i + 1, target - num[i]);32             list.remove(list.size() - 1);33         }34     }35 }

去重这里我用了最土的算法,对每一组新插入的数组,检查result中是否存在相同的组合,这无疑会做很多无用功,而且每次遍历的时间也很浪费,后来偷懒用了之前的去重方法,在每一次进行dfs之前先检查是否与前驱相同。

 1     private void dfs(int[] num,List<Integer> list,int k,int target){ 2         if(target == 0){ 3             List<Integer> copy = new ArrayList<Integer>(list); 4             result.add(copy); 5             return; 6         } 7         if(k >= num.length || num[k] > target) return; 8         for(int i = k; i < num.length; i++){ 9             if(i > k && num[i] == num[i - 1]) continue;//这里本来写的是i>0,结果毫无疑问的跪了。10             list.add(num[i]);11             dfs(num,list,i + 1, target - num[i]);12             list.remove(list.size() - 1);13         }14     }

后来看了jd童鞋的算法,对代码进行了不一样的优化:

完整代码如下:

 1 public class Solution { 2      List<List<Integer>> result = new ArrayList<List<Integer>>(); 3     public List<List<Integer>> combinationSum2(int[] num, int target) { 4         if(num == null || num.length == 0) return result; 5         Arrays.sort(num); 6         List<Integer> list = new ArrayList<Integer>(); 7         dfs(num,list,0,target); 8         return result; 9     }10     11     private void dfs(int[] num,List<Integer> list,int k,int target){12         if(target == 0){13             List<Integer> copy = new ArrayList<Integer>(list);14             result.add(copy);15             return;16         }17         if(k >= num.length || num[k] > target) return;18         for(int i = k; i < num.length; i++){19             list.add(num[i]);20             dfs(num,list,i + 1, target - num[i]);21             list.remove(list.size() - 1);22             while (i < num.length - 1 && num[i] == num[i + 1]) i++;23         }24     }25 }
while (i < num.length - 1 && num[i] == num[i + 1]) i++;
用的很漂亮,总而言之,对DFS还是不能运用自如。
无论是哪种方法,优化完之后,代码的执行次数毫无疑问会减少很多,更主要的是省去了不必要的每次的查重,虽然时间复杂度是一样的,但是从代码可读性和实际执行时间来看,无疑后者更优