3Sum

描述

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, $a \leq b \leq c$)
• 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)

分析

先排序，然后左右夹逼，复杂度 $O(n^2)$。

这个方法可以推广到k-sum，先排序，然后做k-2次循环，在最内层循环左右夹逼，时间复杂度是 $O(\max\{n \log n, n^{k-1}\})$。

代码

// 3Sum
// 先排序，然后左右夹逼，注意跳过重复的数
// Time Complexity: O(n^2)，Space Complexity: O(1)
class Solution {
    public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
        if (nums.size() < 3) return result;
        sort(nums.begin(), nums.end());
        const int target = 0;

        auto last = nums.end();
        for (auto i = nums.begin(); i < last-2; ++i) {
            if (i > nums.begin() && *i == *(i-1)) continue;
            auto j = i+1;
            auto k = last-1;
            while (j < k) {
                if (*i + *j + *k < target) {
                    ++j;
                    while(*j == *(j - 1) && j < k) ++j;
                } else if (*i + *j + *k > target) {
                    --k;
                    while(*k == *(k + 1) && j < k) --k;
                } else {
                    result.push_back({ *i, *j, *k });
                    ++j;
                    --k;
                    while(*j == *(j - 1) && j < k) ++j;
                    while(*k == *(k + 1) && j < k) --k;
                }
            }
        }
        return result;
    }
};

相关题目

• Two sum
• 3Sum Closest
• 4Sum