标签：
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:
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]
解题思路：
修改上题代码，将DFS宽度设置成2即可，注意使用Set，防止重复，JAVA实现如下：
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);
}
}
结果453 ms，效率略低，因此换掉Set，用一个变量计算每次DFS的宽度，ＪＡＶＡ实现如下：
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
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原文地址：http://www.cnblogs.com/tonyluis/p/4495545.html
