转载：http://blog.csdn.net/li4951/article/details/8693212
leetcode上有好几道关于数组中几个数据和为target的题目。恰好正在看剑指offer中“和为s的两个数组这章”，据此思想，leetcode上的三道题目都被我解决了。总结一下。
1.twoSum:
输入一个递增数组和一个数字s，在数组中查找两个数使得它们的和正好是s。
既然题目中已经提到了“递增数组”,那么肯定不会暴力了。因此肯定有<O(n*n)的办法了。
算法：最初我们找到数组的第一个数字和最后一个数字。当两个数字的和大于输入的数字时，把较大的数字往前移动；当两个数字的和小于数字时，把较小的数字往后移动；当相等时，打完收工。这样扫描的顺序是从数组的两端向数组的中间扫描。
链接：http://zhedahht.blog.163.com/blog/static/2541117420072143251809/
2.threeSum:
Given an
array S of n integers,
are there
elements a, b, c in S such
that a + b + c =
0? Find all unique triplets in the array which gives the sum of zero.
Note:
- Elements in a triplet (a,b,c) must
be in non-descending order.
(ie, a ? b ? c) - The solution set must not contain duplicate triplets.
有了twoSum的启发，threeSum所有做的事，只需加上排序，和一层循环。经测试AC。
1 import java.util.ArrayList;
2 import java.util.Arrays;
3
4 public class ThreeSumSolution2 {
5 private ArrayList<ArrayList<Integer>> list;
6
7 public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
8 list = new ArrayList<ArrayList<Integer>>();
9 Arrays.sort(num);
10
11 int i = 0;
12 for (i = 0; i <= num.length - 3; i++) {
13 if (i != 0 && num[i] == num[i - 1]) {
14 continue;
15 }
16 judgeAndPut(num, i, i + 1, num.length - 1);
17 }
18
19 return list;
20 }
21
22 private void judgeAndPut(int[] num, int i, int p, int q) {
23 while (p < q) {
24 if (num[p] + num[q] < -num[i]) {
25 p++;
26 } else if (num[p] + num[q] > -num[i]){
27 q--;
28 } else if (num[p] + num[q] == -num[i]) {
29 ArrayList<Integer> tmpList = new ArrayList<Integer>();
30 tmpList.add(num[i]);
31 tmpList.add(num[p]);
32 tmpList.add(num[q]);
33 list.add(tmpList);
34 p++;
35 q--;
36 while (p < q && num[p] == num[p - 1]) {
37 p++;
38 }
39 while (p < q && num[q] == num[q + 1]) {
40 q--;
41 }
42 }
43 }
44 }
45
46 public static void main(String[] args) {
47 int num[] = {-1,0,1,2,-1,-4};
48 System.out.println(new ThreeSumSolution2().threeSum(num));
49 }
50 }
3. threeSumClosest:
Given an
array S of n integers,
find three integers in S such
that the sum is closest to a given number, target. Return the sum
of the three integers. You may assume that each input would have exactly one
solution.
思路和threeSum几乎一样，只是查找条件有所变化而已。代码更简洁了。
1 import java.util.Arrays;
2
3 public class ThreeSumClosest {
4 private int closest;
5 private boolean needInit;
6
7 public int threeSumClosest(int[] num, int target) {
8 closest = 0;
9 needInit = true;
10 Arrays.sort(num);
11
12 int i = 0;
13 for (i = 0; i <= num.length - 3; i++) {
14 if (i != 0 && num[i] == num[i - 1]) {
15 continue;
16 }
17 judgeAndPut(num, i, i + 1, num.length - 1, target);
18 }
19 return closest;
20 }
21
22 private void judgeAndPut(int[] num, int i, int p, int q, int target) {
23
24 while (p < q) {
25 int sum = num[i] + num[p] + num[q];
26 if (needInit || Math.abs(sum - target) < Math.abs(closest - target)) {
27 closest = sum;
28 }
29 needInit = false;
30
31 if (sum <= target) {
32 p++;
33 while (p < q && num[p] == num[p - 1]) {
34 p++;
35 }
36 } else if (sum > target){
37 q--;
38 while (p < q && num[q] == num[q + 1]) {
39 q--;
40 }
41 }
42 }
43
44 }
45
46 public static void main(String[] args) {
47 int num[] = {0,1,2};
48 System.out.println(new ThreeSumClosest().threeSumClosest(num, 3));
49 }
50 }
4.
fourSum:
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.
再多两层循环，此时已经是n立方的时间效率了，尝试了一下，居然也AC了。不知道还有没有更高效的算法。
1 import java.util.ArrayList;
2 import java.util.Arrays;
3
4 public class FourSum {
5 private ArrayList<ArrayList<Integer>> list;
6
7 public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
8 list = new ArrayList<ArrayList<Integer>>();
9 Arrays.sort(num);
10
11 int i = 0;
12 int j = 0;
13 for (i = 0; i <= num.length - 4; i++) {
14 if (i != 0 && num[i] == num[i - 1]) {
15 continue;
16 }
17
18 for (j = i + 1; j <= num.length - 3; j++) {
19 if (j != i + 1 && num[j] == num[j - 1]) {
20 continue;
21 }
22 judgeAndPut(num, i, j, j + 1, num.length - 1, target);
23 }
24
25 }
26
27 return list;
28 }
29
30 private void judgeAndPut(int[] num, int i, int j, int p, int q, int target) {
31 while (p < q) {
32 int sum = num[i] + num[j] + num[p] + num[q];
33 if (sum < target) {
34 p++;
35 } else if (sum > target){
36 q--;
37 } else if (sum == target) {
38 ArrayList<Integer> tmpList = new ArrayList<Integer>();
39 tmpList.add(num[i]);
40 tmpList.add(num[j]);
41 tmpList.add(num[p]);
42 tmpList.add(num[q]);
43 list.add(tmpList);
44 p++;
45 q--;
46 while (p < q && num[p] == num[p - 1]) {
47 p++;
48 }
49 while (p < q && num[q] == num[q + 1]) {
50 q--;
51 }
52 }
53 }
54
55 }
56
57 public static void main(String[] args) {
58 int num[] = {-1,0,1,0,-2,2};
59 System.out.println(new FourSum().fourSum(num, 0));
60
61 }
62
63 }
转载：http://blog.csdn.net/li4951/article/details/8693212