3Sum 题解
原创文章，拒绝转载
题目来源：https://leetcode.com/problems/3sum/description/
Description
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.
Example
For example, given array S = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
Solution
class Solution {
private:
vector<vector<int> > twoSum(vector<int>& nums, int end, int target) {
vector<vector<int> > res;
int low = 0;
int high = end;
while (low < high) {
if (nums[low] + nums[high] == target) {
vector<int> sum(2);
sum[0] = nums[low++];
sum[1] = nums[high--];
res.push_back(sum);
// 去重
while (low < high && nums[low] == nums[low - 1])
low++;
while (low < high && nums[high] == nums[high + 1])
high--;
} else if (nums[low] + nums[high] > target) {
high--;
} else {
low++;
}
}
return res;
}
public:
vector<vector<int> > threeSum(vector<int>& nums) {
vector<vector<int>> res;
int size = nums.size();
if (size < 3)
return res;
sort(nums.begin(), nums.end());
for (int i = size - 1; i >= 2; i--) {
if (i < size - 1 && nums[i] == nums[i + 1]) // 去重
continue;
auto sum2 = twoSum(nums, i - 1, 0 - nums[i]);
if (!sum2.empty()) {
for (auto& sum : sum2) {
sum.push_back(nums[i]);
res.push_back(sum);
}
}
}
return res;
}
};解题描述
这道题是Two Sum的进阶，解法上采用的是先求Two Sum再根据求到的sum再求三个数和为0的第三个数，不过题意要求不一样，Two Sum要求返回数组下标，这道题要求返回具体的数组元素。而如果使用与Two Sum相同的哈希法去做会比较麻烦。
这里求符合要求的2数之和用的方法是，先将数组排序之后再进行夹逼的办法。并且为了去重，需要在2sum和3sum都进行去重。