题目：
1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
15.3Sum:
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.
For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
18. 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.
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：
这三道题本质上属于同一问题：给定一个数组和目标target，求k个元素，使得k个元素相加的和为target。可能出现的变式为：1.求元素的下标；2.不得重复使用同一元素等，下面进行分析。
对于2Sum而言，要求a+b=target，也就是任意选定元素a，寻找数组中是否有元素b使得b=target-a。可以选择方法有：
方法一：枚举所有的2-subset, 那么这样的复杂度就是从N选出2个，复杂度是O(N^2)
方法二：对数组进行排序并利用头尾两个指针找到两个数使得他们的和等于target。
对于3Sum而言，要求a+b+c=target，这道题可以转化为2Sum问题。即任意选定元素a，在剩余的元素中查找是否存在2个数使得b+c=targe-a。
对于4Sum而言，也可以转化为2Sum问题。任意选定元素a和b，在剩余的元素中查找是否存在2个数使得c+d=targe-a-b。
代码：
2Sum:
1 class Solution {
2 public:
3 vector<int> twoSum(vector<int>& nums, int target) {
4 int l = 0;
5 int r = nums.size() - 1;
6 int i = 0;
7 vector<int> result;
8 multimap<int, int> m;
9 multimap<int, int>::iterator itmulti;
10 for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++) {
11 int temp = *it;
12 m.insert(make_pair(temp, i++));
13 }
14 sort(nums.begin(), nums.end());
15 while (l < r) {
16 if (nums[l] + nums[r] == target) {
17 if (nums[l] == nums[r]) {
18 for (itmulti = m.equal_range(nums[l]).first;
19 itmulti != m.equal_range(nums[l]).second;
20 itmulti++) {
21 result.push_back((*itmulti).second);
22 }
23 } else {
24 itmulti = m.equal_range(nums[l]).first;
25 result.push_back((*itmulti).second);
26 itmulti = m.equal_range(nums[r]).first;
27 result.push_back((*itmulti).second);
28 }
29 break;
30 } else if (nums[l] + nums[r] < target) {
31 l++;
32 } else {
33 r--;
34 }
35 }
36 return result;
37 }
38 };
3Sum:
1 class Solution {
2 public:
3 vector<vector<int> > threeSum(vector<int>& nums) {
4 sort(nums.begin(), nums.end());
5 vector<vector<int> > results;
6 for (int i = 0; i < (signed) nums.size() - 2; i++) {
7 if (i > 0 && nums[i] == nums[i - 1])
8 continue;
9
10 int l = i + 1;
11 int r = nums.size() - 1;
12 while (l < r) {
13 if (nums[l] + nums[r] + nums[i] < 0) {
14 l++;
15 } else if (nums[l] + nums[r] + nums[i] > 0) {
16 r--;
17 } else {
18 vector<int> result;
19 result.push_back(nums[i]);
20 result.push_back(nums[l]);
21 result.push_back(nums[r]);
22 results.push_back(result);
23 while (l < r && nums[l] == nums[l + 1]) {
24 l++;
25 }
26 while (l < r && nums[r] == nums[r - 1]) {
27 r--;
28 }
29 l++;
30 r--;
31 }
32 }
33 }
34 return results;
35 }
36 };
4Sum:
1 class Solution {
2 public:
3 vector<int> twoSum(vector<int>& nums, int target) {
4 int l = 0;
5 int r = nums.size() - 1;
6 int i = 0;
7 vector<int> result;
8 multimap<int, int> m;
9 multimap<int, int>::iterator itmulti;
10 for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++) {
11 int temp = *it;
12 m.insert(make_pair(temp, i++));
13 }
14 sort(nums.begin(), nums.end());
15 while (l < r) {
16 if (nums[l] + nums[r] == target) {
17 if (nums[l] == nums[r]) {
18 for (itmulti = m.equal_range(nums[l]).first;
19 itmulti != m.equal_range(nums[l]).second;
20 itmulti++) {
21 result.push_back((*itmulti).second);
22 }
23 } else {
24 itmulti = m.equal_range(nums[l]).first;
25 result.push_back((*itmulti).second);
26 itmulti = m.equal_range(nums[r]).first;
27 result.push_back((*itmulti).second);
28 }
29 break;
30 } else if (nums[l] + nums[r] < target) {
31 l++;
32 } else {
33 r--;
34 }
35 }
36 return result;
37 }
38 };