Leetcode---3Sum-优快云博客

本文介绍了一种高效算法来寻找整数数组中三个元素之和为零的唯一三元组，强调了排序和双指针技巧的应用，并提供了解决方案的详细步骤。

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:

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)

Have you been asked this question in an interview?

我的解法是2Sum升级版，先按住一个数，然后做剩下的其他数的2Sum。注意之前按住的数之后可以不用考虑。复杂度为n*n。

因为要求所有的组合，所以 num [ i ] + num [ j ] > sum时也要继续往下走，而且要略过所有相同的数。

vector<vector<int> > twoSum(vector<int> &num, int sum){
vector<vector<int>> result;
int i=0;
int j=num.size()-1;
while(i<j){
if(num[i]+num[j]<sum){
i++;
}
else if(num[i]+num[j]>sum){
j--;
}
else{
vector<int> tmp;
tmp.push_back(num[i]);
tmp.push_back(num[j]);
result.push_back(tmp);
int curi=num[i];
int curj=num[j];
while(num[++i]==curi);
while(num[--j]==curj);
}
}
return result;
}
vector<vector<int> > threeSum(vector<int> &num){
vector<vector<int>> result;
if(num.size()<3)
return result;
sort(num.begin(),num.end());
for(int i=0;i<num.size();){
vector<int> twosuminput(num.begin()+i+1,num.end());
vector<vector<int>> twosumresult=twoSum(twosuminput,-num[i]);
for(int j=0;j<twosumresult.size();j++){
vector<int> tmp;
tmp.push_back(num[i]);
tmp.insert(tmp.begin()+1,twosumresult[j].begin(),twosumresult[j].end());
result.push_back(tmp);
}
int cur=num[i];
while(num[++i]==cur);
}
return result;
}

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