原題如下:
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)經過昨天的搜狗面試,我更加註意算法的時間複雜度問題,顯然這道題存在暴力解法,其複雜度爲O(n*n*n),現在考慮的是降低複雜度的問題,首先進行排序,其時間複雜度爲O(nlogn),然後當爲2Sum問題時可以採用左右指針查找的方法(這裏因爲數組中存在重複元素,所以需要去重),此時查找複雜度爲O(n),所以總的時間複雜度爲O(nlogn),在求解3Sum問題時,首先選中一個元素,然後在剩餘的元素中進行2Sum的查找,此時總的時間複雜度爲O(n*n),這裏需要注意的仍是去重問題,首先選擇第一個元素時需要去重,接下來在左右指針移動過程中仍要去重。
class Solution {
public:
vector<vector<int> > threeSum(vector<int> &num) {
sort(num.begin(),num.end());
vector<vector<int>>vv;
if(num.size() < 3)
return vv;
for(int i = 0; i < num.size() - 2; i++){
if(i > 0 && num[i] == num[i - 1])
continue;
twoSum(vv,num,i + 1,num[i] * -1);
}
return vv;
}
void twoSum(vector<vector<int>>&vv,vector<int>num,int index,int target){
int left = index,right = num.size() - 1;
while(left < right){
int temp = num[left] + num[right];
if(temp == target ){
vector<int>triplet;
triplet.push_back(target * -1);
triplet.push_back(num[left]);
triplet.push_back(num[right]);
vv.push_back(triplet);
left++;
right--;
while ( left < num.size() && num[left] == num[left - 1] )
left++;
while( right >= 0 && num[right] == num[right + 1])
right--;
}
else if(temp > target){
right--;
}
else
left++;
}
}
};