Java for LeetCode 040 Combination Sum II

標籤：

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

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]

解題思路：

修改上題代碼，將DFS寬度設定成2即可，注意使用Set，防止重複，JAVA實現如下：

public List<List<Integer>> combinationSum2(int[] candidates, int target) {Set<List<Integer>> list = new HashSet<List<Integer>>();Arrays.sort(candidates);dfs(list, candidates, 0, target, 0);return new ArrayList<List<Integer>>(list);}static List<Integer> list2 = new ArrayList<Integer>();static void dfs(Set<List<Integer>> list, int[] array, int result,int target, int depth) {if (result == target) {list.add(new ArrayList<Integer>(list2));return;}else if (depth >= array.length || result > target)return;for (int i = 0; i <= 1; i++) {for (int j = 0; j < i; j++)list2.add(array[depth]);dfs(list, array, result + array[depth] * i, target, depth+1);for (int j = 0; j < i; j++)list2.remove(list2.size() - 1);}}

 結果453 ms，效率略低，因此換掉Set，用一個變數計算每次DFS的寬度，ＪＡＶＡ實現如下：

public List<List<Integer>> combinationSum2(int[] candidates, int target) {ArrayList<List<Integer>> list = new ArrayList<List<Integer>>();Arrays.sort(candidates);dfs(list, candidates, 0, target, 0);return list;}static List<Integer> list2 = new ArrayList<Integer>();static void dfs(ArrayList<List<Integer>> list, int[] array, int result,int target, int depth) {if (result == target) {list.add(new ArrayList<Integer>(list2));return;}else if (depth >= array.length || result > target)return;int step=1;while(depth<array.length-1&&array[depth]==array[depth+1]){depth++;step++;}for (int i = 0; i <= step; i++) {for (int j = 0; j < i; j++)list2.add(array[depth]);dfs(list, array, result + array[depth] * i, target, depth+1);for (int j = 0; j < i; j++)list2.remove(list2.size() - 1);}}

 

Java for LeetCode 040 Combination Sum II