58

2017.9.25

四数之和其实可以转化为三数之和。

对于数组中的每一个数，可以转化为 求解  threeSum（num[i], target - num[i]）

public class Solution {    /*     * @param numbers: Give an array     * @param target: An integer     * @return: Find all unique quadruplets in the array which gives the sum of zero     */public static List<List<Integer>> fourSum(int[] numbers, int target) {        // write your code hereList<List<Integer>> res = new LinkedList<>();int length = numbers.length;if(length <= 3){return res;}Arrays.sort(numbers);for(int i = 0 ; i < length ; i++){int []num = new int[length-1-i];num = Arrays.copyOfRange(numbers, i+1, length);List<List<Integer>> list = new LinkedList<>();list = threeSum(num,target-numbers[i]);for(int k = 0; k < list.size();k++){List<Integer> tmp = new LinkedList<>();List<Integer> tmp1 = list.get(k);tmp.add(numbers[i]);tmp.addAll(tmp1);res.add(tmp);}while(i < length-1 && numbers[i] == numbers[i+1]){i++;}}return res;    }public static List<List<Integer>> threeSum(int[] numbers, int target) {        // write your code hereList<List<Integer>> res = new LinkedList<>();Arrays.sort(numbers);int length = numbers.length;if(length <=2){return res;}for(int i = 0 ; i < length;i++){// 这里是为了跳过重复的元素for(int j = i+1; j < length; j++){for(int h = j+1; h < length; h++){if(h != length -1 && numbers[h] == numbers[h+1]){continue;}if(numbers[j] + numbers[h] + numbers[i] == target){LinkedList<Integer> list = new LinkedList<Integer>();list.add(numbers[i]);list.add(numbers[j]);list.add(numbers[h]);if(res.contains(list)){continue;}res.add(list);}while(h != length -1 && numbers[h] == numbers[h+1]){h++;}}while(j != length -1 && numbers[j] == numbers[j+1]){j++;}}while(i != length-1 && numbers[i] == numbers[i+1]){i++;}}return res;    }}