LeetCode 遍历技巧 | 18. 4Sum

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/* * Leetcode18. 4Sum * Funtion: 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. * Example:    Input: given array S = [1, 0, -1, 0, -2, 2], and target = 0.    Output: [              [-1,  0, 0, 1],              [-2, -1, 1, 2],              [-2,  0, 0, 2]            ]    Input1:[-3,-2,-1,0,0,1,2,3]  0    Expected1:[[-3,-2,2,3],[-3,-1,1,3],[-3,0,0,3],[-3,0,1,2],[-2,-1,0,3],[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]    Input2:[-1,2,2,-5,0,-1,4]   3    Expected2:[[-5,2,2,4],[-1,0,2,2]] * Author: LKJ * Date:2016/7/29 * Hint:related to 15.3Sum        （穷举第一个和第二个元素，对第三个和第四个元素使用首尾逼近来枚举）        具体思路和求3Sum的O(n2)方法相同，只是去重方法不同        符合条件的元组已经找出了，下次哈希到相同的结果认为它们原来的元组相同（假设哈希未出现冲突），则跳过保存该结果。        为了在求得符合条件的四元组后查找其哈希值时，提高查找速度，使用set或map来存储哈希值，其中的哈希值有序排列        调用这两种容器自身的查找函数可以提高查找效率（内部一般使用二分查找法）。        这里使用map和set在提交结果的时间上有所差别，应该二者内部实现还是有些差别，虽然可以将set也看做一种特殊map。*/#include <iostream>#include <vector>#include <string>#include <cmath>#include <algorithm>using namespace std;class Solution {private:    int myabs(int num){        if(num < 0) num = -num;        return num;    }public:    vector< vector<int> > fourSum(vector<int>& nums, int target) {        sort(nums.begin(),nums.end());        int len = nums.size();        int leftnum;        int rightnum;        int sum;        vector<int> preAns;        vector< vector<int> > ans;        for(int i = 0; i < len-3; i++){            if((i > 0) && (nums[i] == nums[i-1])) continue;//去重            for(int j = i+1; j < len-2; j++){                if((j > i+1) && (nums[j] == nums[j-1])) continue;//去重                leftnum = j+1;                rightnum = len-1;                while(leftnum < rightnum){                    if((leftnum > j+1) && (nums[leftnum] == nums[leftnum-1])) {leftnum++; continue;} //去重                    sum = nums[i] + nums[j] + nums[leftnum] + nums[rightnum];                    if(sum == target){                        preAns.push_back(nums[i]);                        preAns.push_back(nums[j]);                        preAns.push_back(nums[leftnum]);                        preAns.push_back(nums[rightnum]);                        ans.push_back(preAns);                        preAns.clear();                    }                    if(sum > target){                        rightnum--;                        if(nums[rightnum] == nums[rightnum+1]) rightnum--;//去重                    }else{                        leftnum++;                        if(nums[leftnum] == nums[leftnum-1]) leftnum++;//去重                    }                }            }        }        //也可以使用set去重        /*        set< vector<int> > ans;        vector< vector<int> > newans;        set<vector<int>>::iterator it = ans.begin();        for(; it != ans.end(); it++)            newans.push_back(*it);        return newans;        */        return ans;    }};int main(){    int myarr[7] = {-1,2,2,-5,0,-1,4};    vector<int> myin1(myarr,myarr+7);    int myin;    vector< vector<int> > myout;    Solution SA;    cout << "Please Enter" << endl;    cin >> myin;    myout = SA.fourSum(myin1,myin);    cout<<"VecOUT:"<<myout.size()<<endl;    for(unsigned int i = 0; i < myout.size(); i++){        for(vector<int>::iterator itt = myout[i].begin() ;itt!=myout[i].end(); itt++){            cout<<*itt<<"     ";        }        cout<<endl;    }    cout<<endl;    cout<<endl;    return 0;}
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