4Sum

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)

思想：和3sum一致，将n^4转化为n^3
注意事项：end和后一项相等，则向前递归
java代码：

1. ArrayList<ArrayList<Integer> > res = new ArrayList<ArrayList<Integer>>();
2. public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
3. ArrayList<Integer> cur = new ArrayList<Integer>();
4. Arrays.sort(num);
5. int len = num.length;
6. if(len<4) return res;
7. for(int i=0;i<len-3;i++) {
8. if(i>0 && num[i] == num[i-1]) continue;
9. for(int j=i+1;j<len-2;j++) {
10. if(j>i+1 && num[j] == num[j-1]) continue;
11. int sum = target - num[i] - num[j];
12. int start = j+1;
13. int end = len-1;
14. while(start < end) {
15. if(num[start] + num[end] == sum) {
16. cur.add(num[i]);
17. cur.add(num[j]);
18. cur.add(num[start]);
19. cur.add(num[end]);
20. res.add(new ArrayList<Integer>(cur));
21. start++;
22. while(start<end&&num[start]==num[start-1]) start++;
23. end--;
24. while(end>start&&num[end]==num[end+1]) end--;  //不要写成num[end] == num[end-1]
25. cur.clear();
26. } else if(num[start] + num[end] > sum) {
27. end--;
28. while(end>start&&num[end]==num[end+1]) end--;
29. } else {
30. start++;
31. while(start<end&&num[start]==num[start-1]) start++;
32. }
33. }
34. }
35. }
36. return res;
37. }