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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
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