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.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.
【问题分析】
kSUM系列的问题有好多个，如下：
我们对这几个题目分别分析并进行总结。
【思路】
1. Two Sum
解决这个问题可以直接利用两层循环对数组进行遍历，这样的时间复杂度为O(N2)。一个巧妙的办法是利用java中的HashMap来解决这个问题，代码如下：
1 public class Solution {
2 public int[] twoSum(int[] nums, int target) {
3 int[] result = new int[2];
4 Map<Integer, Integer> map = new HashMap<Integer, Integer>();
5 for (int i = 0; i < nums.length; i++) {
6 if (map.containsKey(target - nums[i])) {
7 result[1] = i;
8 result[0] = map.get(target - nums[i]);
9 return result;
10 }
11 map.put(nums[i], i);
12 }
13 return result;
14 }
15 }
由于HashMap的查询效率很高，HashMap的一些操作技巧：http://jiangzhenghua.iteye.com/blog/1196391
2. Two Sum II - Input array is sorted
这个two sum问题中，数组中的元素是已经排序的，我们从数组的头和尾向数组中间靠拢，如果头尾元素相加大于target，则尾指针向前移动一步，如果小于target，则头指针向后移动一步，直到两指针相遇或者相加结果为target。示例如下：[2,3,4,5] target = 7
思路很简单，代码如下：
1 public class Solution {
2 public int[] twoSum(int[] numbers, int target) {
3 int[] result = new int[2];
4 if(numbers == null || numbers.length < 2) return result;
5
6 int left = 0, right = numbers.length-1;
7 while(left < right){
8 int cur = numbers[left] + numbers[right];
9 if(cur == target){
10 result[0] = left+1;
11 result[1] = right+1;
12 return result;
13 }
14 else if(cur < target){
15 left++;
16 }
17 else{
18 right--;
19 }
20 }
21 return result;
22 }
23 }
可见，排序后的two sum效率还是很高的。
3. Three sum
The idea is to sort an input array and then run through all indices of a possible first element of a triplet. For each possible first element we make a standard bi-directional 2Sum sweep of the remaining part of the array. Also we want to skip equal elements to avoid duplicates in the answer without making a set or smth like that.
结合two sum和Two Sum II - Input array is sorted我们可以比较好解决这个问题。上面这段话的思路是：先对数组进行排序，然后遍历排序后数组，把每一个元素当做三元组的开始元素，剩下的两个元素的查找和Two Sum II是相同的。在这个过程中需要注意的就是去重，一些重复出现的元素要跳过。代码如下：
1 public class Solution {
2 public List<List<Integer>> threeSum(int[] nums) {
3 Arrays.sort(nums);
4 List<List<Integer>> result = new LinkedList<>();
5 for(int i = 0; i < nums.length-2; i++){
6 if(i == 0 || (i>0 && nums[i] != nums[i-1])){
7 int target = 0 - nums[i];
8 int left = i + 1;
9 int right = nums.length - 1;
10 while(left < right){
11 if(nums[left] + nums[right] == target){
12 result.add(Arrays.asList(nums[i], nums[left], nums[right]));
13 while(left < right && nums[left] == nums[left+1]) left++;
14 while(left < right && nums[right] == nums[right-1]) right--;
15 left++; right--;
16 }
17 else if (nums[left] + nums[right] < target)
18 left ++;
19 else right--;
20 }
21 }
22 }
23 return result;
24 }
25 }
4. 3Sum Closest
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.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
这个题目在3sum的基础上做了一点变化，要在所有2元组中找到与目标值最接近的三元组的和。我的思路和上一个题目类似，先对数组进行排序，然后在遍历过程中如果发现了更接近的元组的和，则更新最接近的值。如果发现了和值有和目标值相等的，则直接返回目标值。代码如下：
1 public class Solution {
2 public int threeSumClosest(int[] nums, int target) {
3 Arrays.sort(nums);
4 int closest = nums[0]+nums[1]+nums[2];
5 for(int i = 0; i < nums.length-2; i++){
6 int left = i+1, right = nums.length-1;
7 while(left < right){
8 int cur = nums[i] + nums[left] + nums[right];
9 if(cur == target) return cur;
10 else if(cur > target) right--;
11 else left++;
12 if(Math.abs(cur-target) < Math.abs(closest-target))
13 closest = cur;
14 }
15 }
16 return closest;
17 }
18 }
5. 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: 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] ]
这个问题的解决可以借鉴3 sum的思路，只要在3sum外层再增加一层循环即可，代码如下：
1 public class Solution {
2 public List<List<Integer>> fourSum(int[] nums, int target) {
3 Arrays.sort(nums);
4 List<List<Integer>> result = new LinkedList<>();
5 for(int i = 0; i < nums.length-3; i++){
6 if(i == 0 || (i>0 && nums[i] != nums[i-1])){
7 int curtarget1 = target - nums[i];
8 for(int j = i+1; j < nums.length-2; j++){
9 if(j == i+1 || (j>i+1 && nums[j] != nums[j-1])){
10 int curtarget2 = curtarget1 - nums[j];
11 int left = j + 1;
12 int right = nums.length - 1;
13 while(left < right){
14 if(nums[left] + nums[right] == curtarget2){
15 result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
16 while(left < right && nums[left] == nums[left+1]) left++;
17 while(left < right && nums[right] == nums[right-1]) right--;
18 left++; right--;
19 }
20 else if (nums[left] + nums[right] < curtarget2)
21 left ++;
22 else right--;
23 }
24 }
25 }
26 }
27 }
28 return result;
29 }
30 }
自此，这几个解锁的N sum的题目就做完了，这种题目用回溯法适合不适合呢？
另外需要注意在求解的时候要去掉重复的解，如果排序后的元素是a,b,c,d，求解过程如果选定的元素和上一个选定的元素是相同的，则可以直接跳过该元素。至于为什么是这样，大家可以思考一下。
来源：https://www.cnblogs.com/liujinhong/p/5895149.html