标签：collection positive including solution contain
Total Accepted: 34820 Total Submissions: 138493My Submissions
Question Solution
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]
使用堆栈做的，超级不好的方法，没办法啊争取改进
public class Solution {
List<List<Integer>> x=new ArrayList<List<Integer>>();
Map<String, Integer> m=new HashMap<String, Integer>();
void output(Stack<Node> g){
Stack<Node> s=new Stack<Node>();
int[] data=new int[g.size()];
int i=0;
List<Integer> v=new ArrayList<Integer>();
while(!g.isEmpty()){
Node d=g.peek();
//v.add(d.val);
data[i]=d.val;
g.pop();
s.push(d);
i++;
}
while(!s.isEmpty()){
Node d=s.peek();
s.pop();
g.push(d);
}
for(int j=1;j<g.size();j++)
for(int jj=j;jj>0;jj--)
{
if(data[jj]<data[jj-1])
{
int mid=data[jj];
data[jj]=data[jj-1];
data[jj-1]=mid;
}
}
String str=Integer.toString(data[0]);
v.add(data[0]);
for(int j=1;j<g.size();j++)
{
v.add(data[j]);
str=str+" "+Integer.toString(data[j]);
}
if(m.get(str)==null)
{
x.add(v);
m.put(str,1);
}
}
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Stack<Node> stack=new Stack<Node>();
int l=candidates.length;
int i=0;
while(i<l)
{
if(candidates[i]==target)
{
//List<Integer> v=new ArrayList<Integer>();
//v.add(candidates[i]);
//x.add(v);
Node n=new Node(candidates[i], i);
stack.push(n);
output(stack);
i++;
stack.pop();
}
else if(candidates[i]>target)
{
i++;
}
else
{
break;
}
}
if(i==l)
return x;
Node z=new Node(candidates[i],i);
stack.push(z);
int sum=target-z.val;
while(!stack.isEmpty()){
Node p=stack.peek();
int c=p.cur;
if(sum==0)
{
output(stack);
Node u=stack.peek();
stack.pop();
sum=sum+u.val;
if(u.cur<l-1)
{
int h=u.cur;
while(h+1<l)
{
if(candidates[h+1]<=sum)
break;
else
h++;
}
if(h+1<l)
{
Node f=new Node(candidates[h+1],h+1);
stack.push(f);
sum=sum-candidates[h+1];
}
else
{
while(!stack.isEmpty())
{
u=stack.peek();
stack.pop();
sum=sum+u.val;
int hh=u.cur;
while(hh+1<l)
{
if(candidates[hh+1]<=sum)
break;
else
hh++;
}
if(hh+1<l)
{
Node f=new Node(candidates[hh+1],hh+1);
stack.push(f);
sum=sum-candidates[hh+1];
break;
}
}
}
}
else
{
while(!stack.isEmpty())
{
u=stack.peek();
stack.pop();
sum=sum+u.val;
int h=u.cur;
while(h+1<l)
{
if(candidates[h+1]<=sum)
break;
else
h++;
}
if(h+1<l)
{
Node f=new Node(candidates[h+1],h+1);
stack.push(f);
sum=sum-candidates[h+1];
break;
}
}
}
}
else
{
while(c+1<l)
{
if(candidates[c+1]<=sum)
break;
else
c++;
}
if(c+1<l)
{
Node f=new Node(candidates[c+1],c+1);
stack.push(f);
sum=sum-candidates[c+1];
}
else
{
while(!stack.isEmpty())
{
Node u=stack.peek();
stack.pop();
sum=sum+u.val;
if(u.cur<l-1)
{
int h=u.cur;
while(h+1<l)
{
if(candidates[h+1]<=sum)
break;
else
h++;
}
if(h+1<l)
{
Node f=new Node(candidates[h+1],h+1);
stack.push(f);
sum=sum-candidates[h+1];
break;
}
}
}
}
}
}
return x;
}
}
class Node{
int val;
int cur;
Node(int c1, int c2){
val=c1;
cur=c2;
}
}
标签：collection positive including solution contain
原文地址：http://7061299.blog.51cto.com/7051299/1654081
