4Sum

题目描述：

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)

思路和前面的2sum,3sum差不多,先排序，关键就是最后的要把相同的全部都去掉。

这里的indexOf源码：

<pre name="code" class="java">public int indexOf(Object o) {
if (o == null) {
for (int i = 0; i < size; i++)
if (elementData[i]==null)
return i;
} else {
for (int i = 0; i < size; i++)
if (o.equals(elementData[i]))
return i;
}
return -1;
}

equals源码：

public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof List))
return false;

ListIterator<E> e1 = listIterator();
ListIterator<?> e2 = ((List<?>) o).listIterator();
while (e1.hasNext() && e2.hasNext()) {
E o1 = e1.next();
Object o2 = e2.next();
if (!(o1==null ? o2==null : o1.equals(o2)))
return false;
}
return !(e1.hasNext() || e2.hasNext());
}所以当list中查找是按照equals方法来查找的，而equals又支持List，只要每个元素相等就返回true。所以可以这样去除List<List<Integer>>中重复的list

代码如下：

public class Solution {
public List> fourSum(int[] nums, int target) {
List> result=new ArrayList>();
if(nums==null)
return result;
if(nums.length<4)
return result;
Arrays.sort(nums);
int n=nums.length;
for(int i=0;i<=n-4;i++) {
for(int j=i+1;j<=n-3;j++){
int sum=target-nums[i]-nums[j];
int start=j+1,end=n-1;
while(start list=new ArrayList();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[start]);
list.add(nums[end]);
result.add(list);
}
if(nums[start]+nums[end]<=sum)
start++;
else
end--;
}
}
}
List> result2 = new ArrayList>();
for (int s = 0; s < result.size(); s++) {
List list = result.get(s);
if (result2.indexOf(list) == -1) {
result2.add(list);
}
}
return result2;
}

public boolean equals(List list1,List list2){
for(int i=0;i