标签：des blog java io strong for ar art div
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
A solution set is:
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
Solution:
public class Solution {
public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
Arrays.sort(num);
HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
for (int i = 0; i < num.length; i++) {
for (int j = i + 1; j < num.length; j++) {
int k = j + 1;
int l = num.length - 1;
while (k < l) {
int sum = num[i] + num[j] + num[k] + num[l];
if (sum > target) {
l--;
} else if (sum < target) {
k++;
} else if (sum == target) {
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(num[i]);
temp.add(num[j]);
temp.add(num[k]);
temp.add(num[l]);
if (!hashSet.contains(temp)) {
hashSet.add(temp);
result.add(temp);
}
k++;
l--;
}
}
}
}
return result;
}
}
C++
class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
// Note: The Solution object is instantiated only once and is reused by each test case.
vector<int> tmp;
vector<vector<int>> res;
if(num.empty()) return res;
sort(num.begin(), num.end());
for(int i=0; i<num.size(); i++)
{
int cur = target - num[i];
for(int j=i+1; j<num.size(); j++)
{
int temp = cur - num[j];
int start = j+1, end = num.size()-1;
while(start<end)
{
if(num[start]+num[end]==temp)
{
tmp.push_back(num[i]);
tmp.push_back(num[j]);
tmp.push_back(num[start]);
tmp.push_back(num[end]);
res.push_back(tmp);
tmp.clear();
start++;
end--;
while(start<end&&num[start]==num[start-1]) start++;
while(start<end&&num[end]==num[end+1]) end--;
}
else if(num[start]+num[end]<temp)
{
start++;
while(start<end&&num[start]==num[start-1]) start++;
}
else
{
end--;
while(start<end&&num[end]==num[end+1]) end--;
}
}
while(j<num.size()&&num[j]==num[j+1]) j++;
}
while(i<num.size()&&num[i]==num[i+1]) i++;
}
return res;
}
};
标签：des blog java io strong for ar art div
原文地址：http://www.cnblogs.com/yeek/p/3933701.html
