題目:
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] ]
題意解析:
這道題通過我們仔細分析題意來看,其實是尋找兩個數的和等於一個特定的值,如果直接開三層循環遍歷的話,會使算法複雜度變高,一個可以採用的建議是先將這個數組排好序,按順序進行選擇。
首先選取一個數nums[i],令start = i + 1, end = length - 1,。然後判斷三個數的和,若和大於0,則令end--,和小於0,則令start++,找到結果後,將其放入輸出結果即可。
這裏還要注意一點,就是去重問題,這裏當有重複的時候,自動將i自增即可。
一種c++的實現如下:
#include<iostream>
#include<vector>
#include<cstdlib>
#include<cstdio>
using namespace std;
class Solution {
public:
vector<vector<int> > threeSum(vector<int>& nums) {
vector<vector<int> > result;
sort(nums);
int n = nums.size();
for(int i = 0; i < n; i++) {
int temp = nums[i];
int start = 1+i;
int end = n-1;
while(start < end) {
if(nums[start]+nums[end] == 0-temp) {
vector<int> one_solution;
one_solution.push_back(temp);
one_solution.push_back(nums[start]);
one_solution.push_back(nums[end]);
sort(one_solution);
result.push_back(one_solution);
while (start < end && nums[start] == one_solution[1]) start++;
while (start < end && nums[end] == one_solution[2]) end--;
} else if (nums[start] + nums[end] > 0 - temp){
end--;
} else {
start++;
}
}
while (i + 1 < nums.size() && nums[i + 1] == nums[i])
i++;
}
return result;
}
void sort(vector<int>& nums) {
int n = nums.size();
for(int i = 0; i < n-1; i++)
for(int j = 0; j < n-i-1; j++) {
int temp;
if(nums[j] > nums[j+1]) {
swap(nums[j], nums[j+1]);
}
}
}
void swap(int &a, int &b) {
int temp;
temp = a;
a = b;
b = temp;
}
};