Leetcode-18 4Sum

Leetcode-18 4Sum
Zookeeper

Problem description: Givenan arraySOfNIntegers, are there elementsA,B,C, AndDInSSuch thatA+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 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: The KSum problem is eventually converted to the basic 2Sum problem. At the same time, eliminate duplicate values.

  Code:

  Java solution:

   

  Public class Solution {public List
     
      
  > FourSum (int [] num, int target) {List
      
       
  > Result = new ArrayList
       
        
  > (); If (num = null | num. length <= 3) return result; // sorts the array first. Arrays. sort (num); int temp_target = 0; // the target value for every 2Sum conversion (int I = 0; I <num. length-3; I ++) {// eliminate the repeated value while (I! = 0) & (I <num. length-3) & (num [I] = num [I-1]) + + I; // you must pay attention to the array boundary judgment after deduplication, avoid if (I <num. length-3) {for (int j = I + 1; j <num. length-2; j ++) {while (j! = I + 1) & (j <num. length-2) & (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 (start <end) {int temp_sum = num [start] + num [end]; if (temp_sum <temp_target) {++ start ;} else if (temp_sum> temp_target) {-- end;} else {List
        
         
  List = new ArrayList <> (); list. add (num [I]); list. add (num [j]); list. add (num [start ++]); list. add (num [end --]); result. add (list); // continue the search and remove 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
     
       > fourSum(vector
      
        &num, int target) {        // Note: The Solution object is instantiated only once.        vector
       
        > res;    int numlen = num.size();if(num.size()<4) return res;sort(num.begin(),num.end());set
        
         > 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(begin < end){int sum = num[i]+ num[j] + num[begin] + num[end];if(sum == target){vector
         
           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
          
           >::iterator it = tmpres.begin();for(; it != tmpres.end(); it++)res.push_back(*it);return res; }};