# 3Sum

Given an array S of n integers, are there elements abc 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)```
```class Solution {
public:
int b_search(vector<int> &n, int b, int e, int x) {
int l = b;
int r = e;
while (l <= r) {
int mid = l + ((r - l ) >> 2);
if (n[mid] == x) {
return mid;
} else if (n[mid] > x) {
r = mid - 1;
} else {
l = mid + 1;
}
}
return -1;
}
/*
*固定其中2点,找第3点,如果找到就添加到结果中
* 每次固定左边的点,从右想走遍历右边的点,查找两个点的之间是否有满足的点
* 注意左边的点要去掉重复的
*/
vector<vector<int>> threeSum(vector<int>& nums) {
int i,j,k;
int size = nums.size();
vector<vector<int>> res;
sort(nums.begin(), nums.end());
for (i=0; i<size-2; i++) {
//过滤掉相同的左边点,防止出现重复的结果
if (i > 0 && nums[i] == nums[i-1]) {
continue;
}
for (j=size-1; j>i+1; j--) {
if (j < size-1 && nums[j] == nums[j+1]) {
continue;
}
int s = b_search(nums, i+1, j-1, 0 - nums[i] - nums[j]);
if (s != -1) {
vector<int> tmp;
tmp.push_back(nums[i]);
tmp.push_back(nums[s]);
tmp.push_back(nums[j]);
res.push_back(tmp);
}
}
}
return res;
}
};```

3Sum

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