LeetCode 15 — 3Sum(三數之和)
阿新 • • 發佈:2019-01-01
Given an array nums of n integers, are there elements a, b, c in nums 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:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
翻譯
給定一個包含 n 個整數的陣列 nums,判斷 nums 中是否存在三個元素 a,b,c ,使得 a + b + c = 0 ?找出所有滿足條件且不重複的三元組。
注意:答案中不可以包含重複的三元組。
例如, 給定陣列 nums = [-1, 0, 1, 2, -1, -4],
滿足要求的三元組集合為:
[
[-1, 0, 1],
[-1, -1, 2]
]
分析
先將陣列排序,再一個數從頭遍歷,剩下的兩個數從首尾靠近,找三數之和為target的,值得注意的是去除重複。
c++實現
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> res;
if (nums.size() < 3)
return res;
vector<int> tmp;
sort(nums.begin(),nums.end());
for (int i = 0; i < nums.size()-2; i++)
{
if (i > 0 && nums[i] == nums[i-1])
continue;
int j = i+1;
int k = nums.size()-1;
while (j < k)
{
if (nums[i]+nums[j]+nums[k] < 0)
j++;
else if (nums[i]+nums[j]+nums[k] > 0)
k--;
else
{
tmp.clear();
tmp.push_back(nums[i]);
tmp.push_back(nums[j]);
tmp.push_back(nums[k]);
res.push_back(tmp);
j++;
k--;
while (j < k && nums[j] == nums[j-1])
j++;
while (j < k && nums[k] == nums[k+1])
k--;
}
}
}
return res;
}
};