[Leetcode] 114. 4sum

题目

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)

解题之法

// O(n^3)
class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &nums, int target) {
        set<vector<int> > res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < int(nums.size() - 3); ++i) {
            for (int j = i + 1; j < int(nums.size() - 2); ++j) {
                int left = j + 1, right = nums.size() - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        vector<int> out;
                        out.push_back(nums[i]);
                        out.push_back(nums[j]);
                        out.push_back(nums[left]);
                        out.push_back(nums[right]);
                        res.insert(out);
                        ++left; --right;
                    } else if (sum < target) ++left;
                    else --right;
                }
            }
        }
        return vector<vector<int> > (res.begin(), res.end());
    }
};

分析

LeetCode中关于数字之和还有其他几道，分别是Two Sum 两数之和，3Sum 三数之和，3Sum Closest 最近三数之和，虽然难度在递增，但是整体的套路都是一样的，在这里为了避免重复项，我们使用了STL中的set，其特点是不能有重复，如果新加入的数在set中原本就存在的话，插入操作就会失败，这样能很好的避免的重复项的存在。此题的O(n^3)解法的思路跟3Sum 三数之和基本没啥区别，就是多加了一层for循环，其他的都一样。