Leetcode-3sum (the sum of three numbers in the array is 0)

Given an array S of n integers, is 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 is 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)

The first thought must be three-layer loop traversal, time complexity O (n^3), but did the second Sum, feel instantaneous solvable, with the method of Hashtable:

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 ARR Aylist<> ();        List.add (Num[i]);        List.add (Num[j]);        List.add (0-num[i]-num[j]);        Lists.add (list);}}        }        return lists;    }

Time complexity should be in O (n^2), but unfortunately, the direct timeout! Looking at test instructions, need to solve in O (n)? After analysis, certainly not, at least also to O (NLGN). Well, it reminds me of the complexity of sorting, so it's better to control the array first.

First on the code:

Public list<list<integer>> threesum (int[] num) {    list<list<integer>> lists = new ArrayList <> ();    Arrays.sort (num);    for (int i = 0; i < num.length-2; i++) {    if (i = = 0 | | (i > 0 && num[i]! = num[i-1])) {    int low = i+1, high = num.length-1, sum = 0-num[i];    while (Low < high) {    if (num[low]+num[high] = = sum) {    lists.add (arrays.aslist (Num[i], Num[low], Num[high])); C8/>while (Low < high && num[low] = = Num[low+1]) {low    + +;    }    while (Low 

First, control a variable a, linear traversal. At this point only need to find b,c meet b+c=-a can. Because it is sorted well, the rest can be found using the idea of two-point search to find B.C. Then solve (note to eliminate the repetitive solution!!!!) ）



Leetcode-3sum (the sum of three numbers in the array is 0)