Leetcode - 3Sum

蛮常见一道题目。

思路：

1：排序，按顺序遍历两个数之和twoSum，

2： 二分查找 （0 - twoSum)看是否存在

这题最容易错的地方是must not contain duplicate triplets，所以遍历的这时候要用一个数字记录最后一个遍历的数字是，避免重复。

#include<iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
vector<vector<int> > threeSum(vector<int> &num) {

vector<vector<int>> combs;
if (num.size() < 3)
return combs;

sort(num.begin(), num.end());
int prei;
for (int i = 0; i < num.size() - 2; i++)
{
if (i > 0 && num[i] == num[prei])
continue;
if (num[i]>0)
break;

int prej;
for (int j = i + 1; j < num.size() - 1; j++)
{
if (j > i + 1 && num[j] == num[prej])
continue;

int twoSum = num[i] + num[j];
int value = 0 - twoSum;

if (value >= num[j + 1] && value <= num.back())
{
bool flag = searchValue(num, j + 1, num.size() - 1, value);
if (flag == true){
vector<int> comb;
comb.push_back(num[i]);
comb.push_back(num[j]);
comb.push_back(value);
combs.push_back(comb);
}
}
prej = j;
}

prei = i;

}

return combs;
}

bool searchValue(vector<int> num, int left, int right, int value)
{
while (left <= right){
int m = (left + right) / 2;

if (num[m] == value)
return true;
else if (num[m] < value)
left = m + 1;
else
right = m - 1;
}
return false;
}
};