Leetcode-18 4Sum



Problem Description: Givenan ArrayS of n integers, is there elements a, b, C, and D i n 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 is in non-descending order. (ie,a ≤ b ≤ c ≤ d)

• The solution set must not Containduplicate quadruplets.

For example, given array S = {1 0-1 0-22}, and target = 0.

A Solution set is:

( -1, 0, 0, 1)

(-2,-1, 1, 2)

( -2, 0, 0, 2)

Problem Analysis:ksum problem, turn down to the basic 2Sum problem, and pay attention to eliminate duplicate value

Code:

Java Solution:

public class Solution {public list<list<integer>> foursum (int[] num, int target) {list<list< integer>> result = new arraylist<list<integer>> (); if (num = = NULL | | num.length <= 3) return result;// The array is sorted first arrays.sort (num), int temp_target = 0;//The target value of each conversion to 2Sum problem for (int i = 0; i < num.length-3; i++) {//de-duplicated value while ((I ! = 0) && (i < num.length-3) && (num[i] = = num[i-1)) + + i;//after eliminating duplicates, be sure to look at the bounds of the array to avoid any cross-border situations if (I < Nu m.length-3) {for (int j = i + 1, J < Num.length-2; J + +) {while ((j! = i + 1) && (J < num.length-2) &AMP;&A mp (Num[j] = = Num[j-1])) + + j;if (J < num.length-2) {temp_target = Target-num[i]-Num[j];int start = j + 1;int end = Num.length-1;while (STA RT < end) {int temp_sum = Num[start] + num[end]; if (Temp_sum < Temp_target) {+ + start;} else if (Temp_sum > Temp_target) {--end;} else{list<integer> List = new arraylist<> (); List.add (Num[i]); List.add (Num[j]); List.add (Num[start+ +]); List.add (num[end--]); Result.add (list);//continue to search forward and eliminate the same value while (Start < end) && (Num[start] = = Num[start -1]) + + start;while ((Start < end) && (num[end] = = num[end + 1]))--End;}}}}}    return result; }}

Class Solution {public:    vector<vector<int> > Foursum (vector<int> &num, int target) {        // Note:the Solution Object is instantiated only once.        vector<vector<int>> Res;    int numlen = Num.size (), if (Num.size () <4) return Res;sort (Num.begin (), Num.end ());set<vector<int>> tmpres;for (int i = 0; i < Numlen; i++) {for (int j = i+1; J < Numlen; J + +) {int begin = J+1;int end = Numlen-1;while (BE Gin < end) {int sum = num[i]+ Num[j] + num[begin] + num[end];if (sum = = target) {vector<int> tmp;tmp.push_back (num[i ]); Tmp.push_back (Num[j]); Tmp.push_back (Num[begin]); Tmp.push_back (Num[end]); Tmpres.insert (TMP); begin++;end--;} else if (sum<target) begin++;elseend--;}}} Set<vector<int>>::iterator it = Tmpres.begin (); for (; It! = Tmpres.end (); it++) Res.push_back (*it); return res;    };

Leetcode-18 4Sum