LeetCode-4Sum_given an array s of n integer,are there element-优快云博客

本文链接：https://blog.youkuaiyun.com/HolyDominat/article/details/38659257

本文探讨了如何寻找数组中四个元素的组合，使它们的和等于特定目标值的问题，并提供了一个C++实现的示例代码。该算法通过先排序再使用双指针技巧来高效地找出所有可能的四元组，确保结果中不含重复项。

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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)

Solution:

Code:

<span style="font-size:14px;">class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        const int length = num.size();
        vector<vector<int> > results;
        sort(num.begin(), num.end());
        for (int i = 0; i < length-3; ++i) {
            for (int j = i+1; j < length-2; ++j) {
                int m = j+1, n = length-1;
                while (m < n) {
                    if (num[i]+num[j]+num[m]+num[n] == target) {
                        vector<int> result;
                        result.push_back(num[i]);
                        result.push_back(num[j]);
                        result.push_back(num[m]);
                        result.push_back(num[n]);
                        results.push_back(result);
                        while (m+1 < n && num[m+1] == num[m]) ++m;
                        ++m;
                        while (n-1 > m && num[n-1] == num[n]) --n;
                        --n;
                    } else if (num[i]+num[j]+num[m]+num[n] < target) {
                        while (m+1 < n && num[m+1] == num[m]) ++m;
                        ++m;
                    } else {
                        while (n-1 > m && num[n-1] == num[n]) --n;
                        --n;
                    }
                }
                while (j+1 < length-2 && num[j+1] == num[j]) ++j;
            }
            while (i+1 < length-3 && num[i+1] == num[i]) ++i;
        }
        return results;
    }
};</span>