lintcode-medium-4 Sum
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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.
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)
也和3 Sum基本一样。
public class Solution { /** * @param numbers : Give an array numbersbers of n integer * @param target : you need to find four elements that‘s sum of target * @return : Find all unique quadruplets in the array which gives the sum of * zero. */ public ArrayList<ArrayList<Integer>> fourSum(int[] numbers, int target) { /* your code */ ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if(numbers == null || numbers.length == 0) return result; Arrays.sort(numbers); for(int i = 0; i < numbers.length - 3; i++){ if(i > 0 && numbers[i] == numbers[i - 1]) continue; for(int j = i + 1; j < numbers.length - 2; j++){ if(j > i + 1 && numbers[j] == numbers[j - 1]) continue; int left = j + 1; int right = numbers.length - 1; while(left < right){ if(numbers[i] + numbers[j] + numbers[left] + numbers[right] == target){ ArrayList<Integer> line = new ArrayList<Integer>(); line.add(numbers[i]); line.add(numbers[j]); line.add(numbers[left]); line.add(numbers[right]); result.add(new ArrayList<Integer>(line)); left++; while(left < right && numbers[left] == numbers[left - 1]) left++; right--; while(left < right && numbers[right] == numbers[right + 1]) right--; } else if(numbers[i] + numbers[j] + numbers[left] + numbers[right] < target){ left++; while(left < right && numbers[left] == numbers[left - 1]) left++; } else{ right--; while(left < right && numbers[right] == numbers[right + 1]) right--; } } } } return result; } }
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