LeetCode-3SUM（数组中三数之和为0）及其 closest

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.

For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:
(-1, 0, 1)
(-1, -1, 2)

第一想法肯定是三层循环遍历的，时间复杂度O(n^3)，但是做过了了 Two Sum ，感觉瞬间可解，随用hashtable的方法：

public List<List<Integer>> threeSum1(int[] num) {
List<List<Integer>> lists = new ArrayList<>();
HashMap<Integer, Integer> hash = new HashMap<>();
for (int i = 0; i < num.length; i++) {
hash.put(num[i], i);
}
for (int i = 0; i < num.length; i++) {
for (int j = i+1; j < num.length; j++) {
if (hash.get(0-num[i]-num[j]) != null && num[i] <= num[j] && num[j] <= 0-num[i]-num[j]) {
List<Integer> list = new ArrayList<>();
list.add(num[i]);
list.add(num[j]);
list.add(0-num[i]-num[j]);
lists.add(list);
}
}
}
return lists;
}

时间复杂度应该在O(n^2)，但是不幸，直接超时!看着题意，需要在O(n)解决么？经分析，肯定不行，至少也要O（NlgN）。好吧，联想起排序的复杂度，所以，看是还是先把数组给排序了比较好控制。

先上代码：

public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> lists = new ArrayList<List<Integer>>();
if (nums == null || nums.length < 3) return lists;
Arrays.sort(nums);
for (int i = 0; i < nums.length-2; i++) {
// avoid first repeat
if (i > 0 && nums[i] == nums[i-1]) continue;
int low = i+1;
int high = nums.length-1;
while (low < high) {
int sum = nums[i] + nums[low] + nums[high];
if (sum > 0) high--;
else if (sum < 0) low++;
else {
lists.add(new ArrayList<Integer>(Arrays.asList(nums[i], nums[low], nums[high])));
//avoid second repeat
while (low < high && nums[low] == nums[low+1]) low++;
low++;
}
}
}
return lists;
}

先控制一个变量a，线性遍历。此时只需找出b,c满足b+c=-a即可。由于是排序好的，剩下的可以用二分查找的思想找出b.c。于是求解（注意要排除重复解！！！！）

3Sum closest就相对好做许多

public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int ret = nums[0]+nums[1]+nums[2];
for (int i = 0; i < nums.length-2; i++) {
if (i > 0 && nums[i] == nums[i-1]) continue;
int low = i+1;
int high = nums.length-1;
while (low < high) {
int sum = nums[i] + nums[low] + nums[high];
if (sum > target) high--;
else if (sum < target) low++;
else return target;
if (Math.abs(ret-target) > Math.abs(sum-target)) ret = sum;
}
}
return ret;
}