sum无非两种情况，用hashtable 或者用 2 pointer。 Hashtable的时候多不需要考虑重复问题，但是2 pointer往往要考虑重复问题。解决重复问题的方法为初始时判断，如果不是i=0或者nums[i] != nums[i-1]，那就跳过。这样保证第一个值不可能是一样的。同理，在loop里面，当target等于和的时候，左右pointer也要先查重。
- 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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
Hashtable 存nums[i], I;
- 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.
Note: 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]
]
几个关键点：
可能会有重复，需要logic来去掉重复。
另一个关键点是本身不sort，需要sort。
Conditions: sort完最左边大于0或者最右边小于0， 直接return。
2 pointer的时候注意不是一有结果就break，也可能left往右right往左还有符合的结果。
while(left<right && nums[left+1] ==nums[left]) left++;
while(left<right && nums[right-1] ==nums[right]) right--;
- 3Sum Smaller
Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.
For example, given nums = [-2, 0, 1, 3], and target = 2.
Return 2. Because there are two triplets which sums are less than 2:
[-2, 0, 1]
[-2, 0, 3]
Follow up:
Could you solve it in O(n2) runtime?
思路跟3 sum是一样的，只是加的时候， res + = right-left, 因为只要三个和小于target， 那right左边到left的所以都是符合的。
- 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).
这个题也是一样的，只是需要每次判断哪个更小，如果更小，更新res为更小的那个。
- 4Sum
跟3 sum完全一样，多加了一层。需要注意的还是不能重复。