篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了15. 3Sum相关的知识,希望对你有一定的参考价值。
15. 3Sum
题目
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.
For example, given array S = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
解析
- 想写出一次能AC的代码真不容易!
- 很多细节问题,和sumtwo不一样的是:这次有重复元素;sumtwo假定没有重复元素,且只需要返回下标值
- 要跳过重复的元素
class Solution_15 {
public:
void twosum(vector<vector<int>>& vecs,vector<int>& nums,int start, int target)
{
vector<int> ans;
int end = nums.size() - 1;
while (start<end)
{
if (nums[start]+nums[end]==target)
{
ans.push_back(-target);
ans.push_back(nums[start]);
ans.push_back(nums[end]);
vecs.push_back(ans);
ans.clear();
//start++; end--; 跳不过重复的元素
while (start<end&&nums[start]==nums[start+1])
{
start++;
}
while (start<end&&nums[end]==nums[end-1])
{
end--;
}
start++; end--;
}else if (nums[start] + nums[end] < target)
{
start++;
}
else
{
end--;
}
}
return;
}
vector<vector<int>> threeSum(vector<int>& nums) { // Time Limit Exceeded
vector<vector<int>> vecs;
if (nums.size() <= 2)
{
return vecs;
}
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size() - 2;++i)
{
if (i>0&&nums[i]==nums[i-1]) //忽略掉有重复元素的值
{
continue;
}
twosum(vecs, nums,i+1, -nums[i]);
}
return vecs; //
}
vector<vector<int>> threeSum1(vector<int>& nums) { // Time Limit Exceeded
vector<vector<int>> vecs;
if (nums.size()<=2)
{
return vecs;
}
unordered_map<int, int> mp;
vector<int> ans;
for (int i = 0; i < nums.size() - 2;++i)
{
mp.clear();
for (int j = i + 1; j < nums.size();++j)
{
auto iter = mp.find(-(nums[i] + nums[j])); //查找主关键字
if (iter!=mp.end())
{
ans.push_back(nums[i]);
ans.push_back(iter->first);
ans.push_back(nums[j]);
sort(ans.begin(), ans.end()); //处理有重复元素
if (find(vecs.begin(),vecs.end(),ans)==vecs.end()) //没有元素才插入操作
{
vecs.push_back(ans);
}
ans.clear();
}
else
{
mp.insert(make_pair(nums[j], j));
}
}
}
return vecs; //
}
};
题目来源
以上是关于15. 3Sum的主要内容,如果未能解决你的问题,请参考以下文章