【LectCode】15. 3Sum

题目：

15. 3Sum

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• Total Accepted: 210369
• Total Submissions: 978070
• Difficulty: Medium
• Contributor: LeetCode

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]]

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解题思路：这道题需要我们找出所有三个数之和为0的组合，于是和两个数之和类似，我们先固定一个数，首先将数组升序排序，然后我们先固定第一个数，然后从小到大和从大到小扫描数组，分别确定第二和第三个数，若他们之和为0，即为所求，若小于0，则第二个数往右扫描，若大于，则第三个数往左扫描一位，同时注意每次扫描都需要跳过相等的数。

解答：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int> > vec;
        sort(nums.begin(),nums.end());
        int n = nums.size() ;
        for(int i = 0;i < n;i ++){
        while(i > 0 && i < n && nums[i] == nums[i -1]){
        i ++;
}
int j = i +1;
int k = n - 1;
while(j < k){
int sum = nums[i] + nums[j] +nums[k];
if(sum == 0){
vector<int> cur(3);
cur[0] = nums[i];
cur[1] = nums[j];
cur[2] = nums[k];
vec.push_back(cur);
j ++;
k --; 
while(j < k && nums[j] == nums[j -1]){
j ++;
}
while(k > j && nums[k] == nums[k+1]){
k --;
}
}
else if(sum < 0){
j ++;
while(j < k && nums[j] == nums[j -1]){
j ++;
}
}
else{
k --;
while(k > j && nums[k] == nums[k+1]){
k --;
}
}
}

}
return vec;
    }
};

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Submission Result: Accepted More Details