4Sum

题目：

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:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.

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)

分析：

2-sum,3-sum,4-sum
http://blog.csdn.net/linhuanmars/article/details/24826871

public class Solution {
public List<List<Integer>> fourSum(int[] num, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(num==null||num.length==0)
return res;
Arrays.sort(num);
for(int i=num.length-1;i>2;i--)
{
if(i==num.length-1 || num[i]!=num[i+1])
{
ArrayList<ArrayList<Integer>> curRes = threeSum(num,i-1,target-num[i]);
for(int j=0;j<curRes.size();j++)
{
curRes.get(j).add(num[i]);
}
res.addAll(curRes);
}
}
return res;
}
private ArrayList<ArrayList<Integer>> threeSum(int[] num, int end, int target)
{
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
for(int i=end;i>1;i--)
{
if(i==end || num[i]!=num[i+1])
{
ArrayList<ArrayList<Integer>> curRes = twoSum(num,i-1,target-num[i]);
for(int j=0;j<curRes.size();j++)
{
curRes.get(j).add(num[i]);
}
res.addAll(curRes);
}
}
return res;
}
private ArrayList<ArrayList<Integer>> twoSum(int[] num, int end, int target)
{
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
int l=0;
int r=end;
while(l<r)
{
if(num[l]+num[r]==target)
{
ArrayList<Integer> item = new ArrayList<Integer>();
item.add(num[l]);
item.add(num[r]);
res.add(item);
l++;
r--;
while(l<r&&num[l]==num[l-1])
l++;
while(l<r&&num[r]==num[r+1])
r--;
}
else if(num[l]+num[r]>target)
{
r--;
}
else
{
l++;
}
}
return res;
}
}

还有一种解法，闲了再研究(*^__^*) ……