[LeetCode]4Sum

最新推荐文章于 2019-01-04 21:54:43 发布

原创最新推荐文章于 2019-01-04 21:54:43 发布 · 334 阅读

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本文介绍了一种高效算法解决四数之和问题，即在给定数组中寻找四个元素之和等于目标值的所有组合。通过将问题简化为三数之和，并在数组中进行遍历优化，实现复杂度降低至O(N^3)，同时避免了重复解的计算。

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)

4sum,可以归纳为3sum问题，这样复杂度 O(N^3)。但是实际中可以通过一些细节处理提高遍历无用解的情形。

16ms

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> ret;
        int len = nums.size();
        sort(nums.begin(),nums.end());
        for(int i=0;i<len-3;++i){
            if(i>0&&nums[i]==nums[i-1]) continue;//dup start
            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;//no result
            if(nums[i]+nums[len-1]+nums[len-2]+nums[len-3]<target) continue;//at this start no result,go to next one,important
            for(int j=i+1;j<len-2;++j){//3 sum
                if(j>i+1&&nums[j]==nums[j-1]) continue;//dup start
                if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;//优化的关键
                if(nums[i]+nums[j]+nums[len-1]+nums[len-2]<target) continue; //优化的关键
                int left = j+1;
                int right = len-1;
                while(left<right){
                    if(nums[i]+nums[j]+nums[left]+nums[right]==target){
                        ret.push_back({nums[i],nums[j],nums[left],nums[right]});
                        ++left;
                        --right;
                        while(nums[left]==nums[left-1]&&nums[right]==nums[right+1]){ //重复子解,重要
                            ++left;
                            --right;
                        }
                    }
                    else if(nums[i]+nums[j]+nums[left]+nums[right]<target){ 
                        ++left;
                        while(nums[left]==nums[left-1]){//重复解，重要
                            ++left;
                        }
                    }
                    else if(nums[i]+nums[j]+nums[left]+nums[right]>target){ 
                        --right;
                        while(nums[right]==nums[right+1]){ //重复解
                            --right; 
                        }
                    }
                }
            }
        }
        return ret;
    }
};