LeetCode(18)-4Sum

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问题描述：

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)

问题分析：

跟3Sum一样，先排序，再首尾夹逼，得到字串。

解题方法：

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> ret;        int len = nums.size();                if( len < 4){            return ret;        }                sort(nums.begin(), nums.end());                for(int i = 0; i <= len-4; i++){            for(int z = len -1; z >=3; z--){                int j = i + 1;                int k = z - 1;                int sum = nums[i] + nums[j] + nums[k] + nums[z];                while(j < k){                    if( sum < target){                        j++;                    }                    else if(sum > target){                        k--;                    }                    else{                        ret.push_back({ nums[i], nums[j], nums[k], nums[z] });                        j++;                        k--;                    }                    while( j < k && nums[j] == nums[j-1] )  j++;                    while(j < k && nums[k] == nums[k+1]) k--;                }                while( z < len -1 && nums[z] == nums[z+1]) z--;            }            while(i > 0 && nums[i] == nums[i-1]) i++;        }        return ret;    }};

这是我写的程序，编译器提示answer wrong，百思不得其解。等以后在看。

下一个是在github上的标准答案：

#include <bits/stdc++.h>using namespace std;const int N = 0;class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        vector<vector<int> > ret;        int len = num.size();        if (len <= 3)            return ret;        sort(num.begin(), num.end());        for (int i = 0; i <= len - 4; i++) {            for (int m = i + 1; m <= len - 3; m++) {                int j = m + 1;                int k = len - 1;                while (j < k) {                    if (num[i] + num[m] + num[j] + num[k] < target) {                        ++j;                    } else if (num[i] + num[m] +  num[j] + num[k] > target) {                        --k;                    } else {                        ret.push_back({ num[i], num[m], num[j], num[k] });                        ++j;                        --k;                        while (j < k && num[j] == num[j - 1])                            ++j;                        while (j < k && num[k] == num[k + 1])                            --k;                    }                }                while (m < len-3 && num[m] == num[m + 1])                    ++m;            }            while (i < len-4 && num[i] == num[i + 1])                ++i;        }        return ret;    }};int main() {    vector<int> num;    int n, tar;    cin >> tar;    while (~scanf("%d", &n))        num.push_back(n);    Solution s;    vector<vector<int> > ret = s.fourSum(num, tar);    for (auto &i : ret) {        for (auto &j : i)            cout << j << ' ';        cout << endl;    }    return 0;}

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