2014
11-18

# LeetCode-3Sum[数组]

### 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:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, abc)
• 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)


// LeetCode, 3Sum
// 先排序，然后左右夹逼，时间复杂度O(n^2)，空间复杂度O(1)
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& num) {
vector<vector<int>> result;
if (num.size() < 3) return result;
sort(num.begin(), num.end());
const int target = 0;

auto last = num.end();
for (auto a = num.begin(); a < prev(last, 2); ++a) {
auto b = next(a);
auto c = prev(last);
while (b < c) {
if (*a + *b + *c < target) {
++b;
} else if (*a + *b + *c > target) {
--c;
} else {
result.push_back({ *a, *b, *c });
++b;
--c;
}
}
}
sort(result.begin(), result.end());
result.erase(unique(result.begin(), result.end()), result.end());
return result;
}
};


Java代码:

public class Solution {
public List<List<Integer>> threeSum(int[] num) {
List<List<Integer>> result = new ArrayList();

Arrays.sort(num);
int start,end,temp;
for(int i=0;i<num.length;i++){
if( i!=0 && num[i]==num[i-1] )continue;     //num 1：only reserve first of all same values
int current=num[i];
start=i+1;
end=num.length-1;

while(start<end){
if(start!=i+1 && num[start]==num[start-1] ){        //num 2：only reserve first of all same values
start++;
continue;
}
temp=num[start]+num[end];

if(temp==-current){                 //find
List<Integer> list = new ArrayList<Integer>(3);
}