*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)
题解： 4 sum跟3 sum是一样的思路，只不过需要多考虑一个加数，这样时间复杂度变为O(n3)。
使用HashSet来解决重复问题的代码如下：

public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
Arrays.sort(num);
for (int i = 0; i <= num.length-4; i++) {
for (int j = i + 1; j <= num.length-3; j++) {
int low = j + 1;
int high = num.length - 1;

while (low < high) {
int sum = num[i] + num[j] + num[low] + num[high];

if (sum > target) {
high--;
} else if (sum < target) {
low++;
} else if (sum == target) {
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(num[i]);
temp.add(num[j]);
temp.add(num[low]);
temp.add(num[high]);

if (!hashSet.contains(temp)) {
hashSet.add(temp);
result.add(temp);
}

low++;
high--;
}
}
}
}

return result;
}

使用挪动指针的方法来解决重复的代码如下：

public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
Arrays.sort(num);
for (int i = 0; i <= num.length-4; i++) {
if(i==0||num[i]!=num[i-1]){
for (int j = i + 1; j <= num.length-3; j++) {
if(j==i+1||num[j]!=num[j-1]){
int low = j + 1;
int high = num.length - 1;

while (low < high) {
int sum = num[i] + num[j] + num[low] + num[high];

if (sum > target) {
high--;
} else if (sum < target) {
low++;
} else if (sum == target) {
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(num[i]);
temp.add(num[j]);
temp.add(num[low]);
temp.add(num[high]);

if (!hashSet.contains(temp)) {
hashSet.add(temp);
result.add(temp);
}

low++;
high--;

while(low<high&&num[low]==num[low-1])//remove dupicate
low++;
while(low<high&&num[high]==num[high+1])//remove dupicate
high--;
}
}
}
}
}
}

return result;
}